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Signal ANALYSIS MODEL OBJECTIVE QUESTIONS

1. The Z-Transform X(z) of a discrete time signal x(n) is defined as:
1a
View Answer
Answer: b
Explanation: The z-transform of a real discrete time sequence x(n) is defined as a power of ‘z’ which is equal to1b, where ‘z’ is a complex variable.
2. What is the set of all values of z for which X(z) attains a finite value?
a) Radius of convergence
b) Radius of divergence
c) Feasible solution
d) None of the mentioned
View Answer
Answer: a
Explanation: Since X(z) is a infinite power series, it is defined only at few values of z. The set of all values of z where X(z) converges to a finite value is called as Radius of Convergence(ROC).
3. What is the z-transform of the finite duration signal3a
3b
View Answer
Answer: d
Explanation: We know that, for a given signal x(n) the z-transform is defined as1b
Substitute the values of n from -2 to 3 and the corresponding signal values in the above formula
We get,3c
4. What is the ROC of the signal x(n)=δ(n-k),k>0?
a) z=0
b) z=∞
c) Entire z-plane, except at z=0
d) Entire z-plane, except at z=∞
View Answer
Answer: c
Explanation: We know that, the z-transform of a signal x(n) is1b
Given x(n)= δ(n-k)=1 at n=k
=> X(z)=z-k
From the above equation, X(z) is defined at all values of z except at z=0 for k>0.
So ROC is defined as Entire z-plane, except at z=0.
5. What is the z-transform of the signal x(n)=(0.5)nu(n)?
5a
View Answer
Answer: a
Explanation: For a given signal x(n), its z-transform1b
5b
6. Which of the following series has an ROC as mentioned below?
6a
a) α-nu(n)
b) αnu(n)
c) α-nu(-n)
d) αnu(n)
View Answer
Answer: b
Explanation:
Let x(n)= αnu(n)
6b
7. What is the z-transform of the signal x(n)= -αnu(-n-1)?
7a
View Answer
Answer: d
Explanation:
7b
8. What is the ROC of the z-transform of the signal x(n)= anu(n)+bnu(-n-1)?
a) |a|<|z|<|b|
b) |a|>|z|>|b|
c) |a|>|z|<|b|
d) |a|<|z|>|b|
View Answer
Answer: a
Explanation: We know that,
ROC of z-transform of a<sup>n</sup>u(n) is |z|>|a|.
ROC of z-transform of b<sup>n</sup>u(-n-1) is |z|<|b|.
By combining both the ROC's we get the ROC of z-transform of the signal x(n) as |a|<|z|<|b|.
9. What is the ROC of z-transform of finite duration anti-causal sequence?
a) z=0
b) z=∞
c) Entire z-plane, except at z=0
d) Entire z-plane, except at z=∞
View Answer
Answer: d
Explanation: Let us an example of anti causal sequence whose z-transform will be in the form X(z)=1+z+z2 which has a finite value at all values of ‘z’ except at z=∞.So, ROC of an anti-causal sequence is entire z-plane except at z=∞.
10. What is the ROC of z-transform of an two sided infinite sequence?
a) |z|>r1
b) |z|<r1
c) r2<|z|<r1
d) None of the mentioned
View Answer
Answer: c
Explanation: Let us plot the graph of z-transform of any two sided sequence which looks as follows.
10
From the above graph, we can state that the ROC of a two sided sequence will be of the form r2 < |z| < r1.
11. The z-transform of a sequence x(n) which is given as1bis known as:
a) Uni-lateral Z-transform
b) Bi-lateral Z-transform
c) Tri-lateral Z-transform
d) None of the mentioned
View Answer
Answer: b
Explanation: The entire timing sequence is divided into two parts n=0 to ∞ and n=-∞ to 0.
Since the z-transform of the signal given in the questions contains both the parts, it is called as Bi-lateral z-transform.
12. What is the ROC of the system function H(z) if the discrete time LTI system is BIBO stable?
a) Entire z-plane, except at z=0
b) Entire z-plane, except at z=∞
c) Contain unit circle
d) None of the mentioned
View Answer
Answer: c
Explanation: A discrete time LTI is BIBO stable, if and only if its impulse response h(n) is absolutely summable. That is,
12
13. The ROC of z-transform of any signal cannot contain poles.
a) True
b) False
View Answer
Answer: a
Explanation: Since the value of z-transform tends to infinity, the ROC of the z-transform does not contain poles.
14. Is the discrete time LTI system with impulse response h(n)=an(n) (|a| < 1) BIBO stable?
a) True
b) False
View Answer
Answer: a
Explanation:
Given h(n)= a<sup>n</sup>(n) (|a|<1)
The z-transform of h(n) is H(z)=z/(z-a),ROC is |z|>|a|
If |a|<1, then the ROC contains the unit circle. So, the system is BIBO stable.
15. What is the ROC of a causal infinite length sequence?
a) |z|<r1
b) |z|>r1
c) r2<|z|<r1
d) None of the mentioned
View Answer
Answer: b
Explanation: The ROC of causal infinite sequence is of form |z|>r1 where r1 is largest magnitude of poles.

1. Which of the following justifies the linearity property of z-transform?[x(n)↔X(z)] a) x(n)+y(n) ↔X(z)Y(z)
b) x(n)+y(n) ↔X(z)+Y(z)
c) x(n)y(n) ↔X(z)+Y(z)
d) x(n)y(n) ↔X(z)Y(z)
View Answer
Answer: b
Explanation: According to the linearity property of z-transform, if X(z) and Y(z) are the z-transforms of x(n) and y(n) respectively then, the z-transform of x(n)+y(n) is X(z)+Y(z).
2. What is the z-transform of the signal x(n)=[3(2n)-4(3n)]u(n)?
a) 3/(1-2z-1)-4/(1-3z-1)
b) 3/(1+2z-1)-4/(1+3z-1)
c) 3/(1-2z)-4/(1-3z)
d) None of the mentioned
View Answer
Answer: a
Explanation: Let us divide the given x(n) into x1(n)= 3(2n)u(n) and x2(n)= 4(3n)u(n)
and x(n)=x1(n)-x2(n)
From the definition of z-transform X1(z)= 3/(1-2z-1) and X2(z)= 4/(1-3z-1)
So, from the linearity property of z-transform
X(z)=X1(z)-X2(z)
=> X(z)= 3/(1-2z-1)-4/(1-3z-1)
3. What is the z-transform of the signal x(n)=sin(jω0n)u(n)?
3a
View Answer
Answer: d
Explanation: By Euler’s identity, the given signal x(n) can be written as
3b
4. According to Time shifting property of z-transform, if X(z) is the z-transform of x(n) then what is the z-transform of x(n-k)?
a) zkX(z)
b) z-kX(z)
c) X(z-k)
d) X(z+k)
View Answer
Answer: b
Explanation: According to the definition of Z-transform
4
5. What is the z-transform of the signal defined as x(n)=u(n)-u(n-N)?
5a
View Answer
Answer: d
Explanation:
5b
6. If X(z) is the z-transform of the signal x(n) then what is the z-transform of anx(n)?
a) X(az)
b) X(az-1)
c) X(a-1z)
d) None of the mentioned
View Answer
Answer: c
Explanation: We know that from the definition of z-transform
6
7. If the ROC of X(z) is r1<|z|<r2, then what is the ROC of X(a-1z)?
a) |a|r1<|z|<|a|r2
b) |a|r1>|z|>|a|r2
c) |a|r1<|z|>|a|r2
d) |a|r1>|z|<|a|r2
View Answer
Answer: a
Explanation: Given ROC of X(z) is r1<|z|<r2
Then ROC of X(a-1z) will be given by r1<|a-1z |<r2=|a|r1<|z|<|a|r2
8. What is the z-transform of the signal x(n)=an(sinω0n)u(n)?
8a
View Answer
Answer: d
Explanation:
8b
9. If X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal x(-n)?
a) X(-z)
b) X(z-1)
c) X-1(z)
d) None of the mentioned
View Answer
Answer: b
Explanation: From the definition of z-transform, we have
9
10. X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal nx(n)?
a) -z(dX(z))/dz
b) zdX(z)/dz
c) -z-1dX(z)/dz
d) z-1(dX(z))/dz
View Answer
Answer: a
Explanation:
From the definition of z-transform, we have
10
11. What is the z-transform of the signal x(n)=nanu(n)?
11a
View Answer
Answer: c
Explanation:
11b
1. Which of the following method is used to find the inverse z-transform of a signal?
a) Counter integration
b) Expansion into a series of terms
c) Partial fraction expansion
d) All of the mentioned
View Answer
Answer: d
Explanation: All the methods mentioned above can be used to calculate the inverse z-transform of the given signal.
2. What is the inverse z-transform of X(z)=1/(1-1.5z-1+0.5z-2 ) if ROC is |z|>1?
a) {1,3/2,7/4,15/8,31/16,….}
a) {1,2/3,4/7,8/15,16/31,….}
a) {1/2,3/4,7/8,15/16,31/32,….}
d) None of the mentioned
View Answer
Answer: a
Explanation: Since the ROC is the exterior circle, we expect x(n) to be a causal signal. Thus we seek a power series expansion in negative powers of ‘z’. By dividing the numerator of X(z) by its denominator, we obtain the power series
2b
So, we obtain x(n)= {1,3/2,7/4,15/8,31/16,….}.
3. What is the inverse z-transform of X(z)=1/(1-1.5z-1+0.5z-2 ) if ROC is |z| < 0.5?
a) {….62,30,14,6,2}
b) {…..62,30,14,6,2,0,0}
c) {0,0,2,6,14,30,62…..}
d) {2,6,14,30,62…..}
View Answer
Answer: b
Explanation: In this case the ROC is the interior of a circle. Consequently, the signal x(n) is anti causal. To obtain a power series expansion in positive powers of z, we perform the long division in the following way:
3a
Thus3b
In this case x(n)=0 for n≥0.Thus we obtain x(n)= {…..62,30,14,6,2,0,0}
4. What is the inverse z-transform of X(z)=log(1+az-1) |z|>|a|?
4ad)None of the mentioned
View Answer
Answer: c
Explanation: Using the power series expansion for log(1+x), with |x|<1, we have4b
5. What is the proper fraction and polynomial form of the improper rational transform
X(z)= (1+3z-1+11/6 z-2+1/3 z-3)/(1+5/6 z-1+1/6 z-2 )?
a) 1+2z -1+(1/6 z-1)/(1+5/6 z-1+1/6 z-2 )
b) 1-2z -1+(1/6 z-1)/(1+5/6 z-1+1/6 z-2 )
c) 1+2z -1+(1/3 z-1)/(1+5/6 z-1+1/6 z-2)
d) 1+2z -1-(1/6 z-1)/(1+5/6 z-1+1/6 z-2 )
View Answer
Answer: a
Explanation: First, we note that we should reduce the numerator so that the terms z-2 and z -3 are eliminated. Thus we should carry out the long division with these two polynomials written in the reverse order. We stop the division when the order of the remainder becomes z -1. Then we obtain
X(z)= 1+2z -1+(1/6 z-1)/(1+5/6 z-1+1/6 z-2 )
6. What is the partial fraction expansion of the proper function X(z)= 1/(1-1.5z-1+0.5z-2 )?
a) 2z/(z-1)-z/(z+0.5)
b) 2z/(z-1)+z/(z-0.5)
c) 2z/(z-1)+z/(z+0.5)
d) 2z/(z-1)-z/(z-0.5)
View Answer
Answer: d
Explanation: First we eliminate the negative powers of z by multiplying both numerator and denominator by z2.
Thus we obtain X(z)= z2/(z2-1.5z+0.5)
The poles of X(z) are p1=1 and p2=0.5. Consequently, the expansion will be
(X(z))/z = z/((z-1)(z-0.5)) = 2/((z-1) ) – 1/((z-0.5) )( obtained by applying partial fractions)
=>X(z)= 2z/(z-1)-z/(z-0.5)
7. What is the partial fraction expansion of X(z)= (1+z-1)/(1-z-1+0.5z-2 )?
a) (z(0.5-1.5j))/(z-0.5-0.5j) – (z(0.5+1.5j))/(z-0.5+0.5j)
b) (z(0.5-1.5j))/(z-0.5-0.5j) + (z(0.5+1.5j))/(z-0.5+0.5j)
c) (z(0.5+1.5j))/(z-0.5-0.5j) – (z(0.5-1.5j))/(z-0.5+0.5j)
d) (z(0.5+1.5j))/(z-0.5-0.5j) + (z(0.5-1.5j))/(z-0.5+0.5j)
View Answer
Answer: b
Explanation: To eliminate the negative powers of z, we multiply both numerator and denominator by z2. Thus,
X(z)=(z(z+1))/(z-2-z+0.5)
The poles of X(z) are complex conjugates p1=0.5+0.5j and p2=0.5-0.5j
Consequently the expansion will be
X(z)= (z(0.5-1.5j))/(z-0.5-0.5j) + (z(0.5+1.5j))/(z-0.5+0.5j)
8. What is the partial fraction expansion of X(z)=1/((1+z-1 )(1-z-1)2)?
a) z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z+1)〗2 )
b) z/(4(z+1)) + 3z/(4(z-1)) – z/(2〖(z+1)〗2 )
c) z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z-1)〗2 )
d) z/(4(z+1)) + z/(4(z-1)) + z/(2〖(z+1)〗2 )
View Answer
Answer: c
Explanation: First we express X(z) in terms of positive powers of z, in the form X(z)=z3/((z+1)〖(z-1)〗2 )
X(z) has a simple pole at z=-1 and a double pole at z=1. In such a case the approximate partial fraction expansion is
(X(z))/z = z2/((z+1)〖(z-1)〗2 ) =A/(z+1) + B/(z-1) + C/〖(z-1)〗2
On simplifying, we get the values of A, B and C as 1/4, 3/4 and 1/2 respectively.
Therefore, we get X(z)= z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z-1)〗2 ) .
9. What is the inverse z-transform of X(z)= 1/(1-1.5z-1+0.5z2-2 ) if ROC is |z|>1?
a) (2-0.5n)u(n)
b) (2+0.5n)u(n)
c) (2n-0.5n)u(n)
d) None of the mentioned
View Answer
Answer: a
Explanation: The partial fraction expansion for the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In case when ROC is |z|>1, the signal x(n) is causal and both the terms in the above equation are causal terms. Thus, when we apply inverse z-transform to the above equation, we get
x(n)=2(1)nu(n)-(0.5)nu(n)=(2-0.5n)u(n)
10. What is the inverse z-transform of X(z)= 1/(1-1.5z-1+0.5z-2 ) if ROC is |z|<0.5? a) [-2-0.5n]u(n)
b) [-2+0.5n]u(n)
c) [-2+0.5n]u(-n-1)
d) [-2-0.5n]u(-n-1)
View Answer
Answer: c
Explanation: The partial fraction expansion for the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In case when ROC is |z|<0.5,the signal is anti causal. Thus both the terms in the above equation are anti causal terms. So, if we apply inverse z-transform to the above equation we get x(n)= [-2+0.5n]u(-n-1)
11. What is the inverse z-transform of X(z)= 1/(1-1.5z-1+0.5z-2 ) if ROC is 0.5<|z|<1? a) -2u(-n-1)+(0.5)nu(n)
b) -2u(-n-1)-(0.5)nu(n)
c) -2u(-n-1)+(0.5)nu(-n-1)
d) 2u(n)+(0.5)nu(-n-1)
View Answer
Answer: b
Explanation: The partial fraction expansion of the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In this case ROC is 0.5<|z|<1 is a ring, which implies that the signal is two sided. Thus one of the signal corresponds to a causal signal and the other corresponds to an anti causal signal. Obviously, the ROC given is the overlapping of the regions |z|>0.5 and |z|<1. Hence the pole p2=0.5 provides the causal part and the pole p1=1 provides the anti causal part. SO, if we apply the inverse z-transform we get x(n)= -2u(-n-1)-(0.5)nu(n)
12.What is the causal signal x(n) having the z-transform X(z)= 1/((1+z-1 ) [(1-z-1)]2 )?
a)[1/4(-1)n+3/4-n/2]u(n)
b)[1/4(-1)n+3/4-n/2]u(-n-1)
c)[1/4+3/4(-1)n-n/2]u(n)
d)[1/4(-1)n+3/4+n/2]u(n)
View Answer
Answer: d
Explanation: The partial fraction expansion of X(z) is X(z)= z/(4(z+1)) + 3z/(4(z-1)) + z/(2[(z-1)]2)
When we apply the inverse z-transform for the above equation, we get
x(n)=[1/4(-1)n+3/4+n/2]u(n)


1. If x(n) is a discrete-time signal, then the value of x(n) at non integer value of ‘n’ is:
a) Zero
b) Positive
c) Negative
d) Not defined
View Answer
Answer: d
Explanation: For a discrete time signal, the value of x(n) exists only at integral values of n. So, for a non- integer value of ‘n’ the value of x(n) does not exist.
2. The discrete time function defined as u(n)=n for n≥0;=0 for n<0 is an:
a) Unit sample signal
b) Unit step signal
c) Unit ramp signal
d) None of the mentioned
View Answer
Answer: c
Explanation: When we plot the graph for the given function, we get a straight line passing through origin with a unit positive slope. So, the function is called as unit ramp signal.
3.The phase function of a discrete time signal x(n)=an, where a=r.ejθ is:
a) tan(nθ)
b) nθ
c) tan-1(nθ)
d) None of the mentioned
View Answer
Answer: b
Explanation: Given x(n)=an=(r.ejθ)n =rn.ejnθ
=>x(n)=rn.(cosnθ+jsinnθ)
Phase function is tan-1(cosnθ/sinnθ)=tan-1(tan nθ)=nθ
4. The signal given by the equation4is known as:
a) Energy signal
b) Power signal
c) Work done signal
d) None of the mentioned
View Answer
Answer: a
Explanation: We have used the magnitude-squared values of x(n), so that our definition applies to complex-valued as well as real-valued signals. If the energy of the signal is finite i.e., 0<E<∞ then the given signal is known as Energy signal.
5. x(n)*δ(n-k)=?
a) x(n)
b) x(k)
c) x(k)*δ(n-k)
d) x(k)*δ(k)
View Answer
Answer: c
Explanation: The given signal is defined only when n=k by the definition of delta function. So, x(n)*δ(n-k)= x(k)*δ(n-k)
6. A real valued signal x(n) is called as anti-symmetric if:
a) x(n)=x(-n)
b) x(n)=-x(-n)
c) x(n)=-x(n)
d) None of the mentioned
View Answer
Answer: b
Explanation: According to the definition of anti-symmetric signal, the signal x(n) should be symmetric over origin. So, for the signal x(n) to be symmetric, it should satisfy the condition x(n)=-x(-n).
7. The odd part of a signal x(t) is:
a) x(t)+x(-t)
b) x(t)-x(-t)
c) (1/2)*(x(t)+x(-t))
d) (1/2)*(x(t)-x(-t))
View Answer
Answer: d
Explanation: Let x(t)=xe(t)+xo(t)
=>x(-t)=xe(-t)-xo(-t)
By subtracting the above two equations, we get
xo(t)=(1/2)*(x(t)-x(-t))
8. Time scaling operation is also known as:
a) Down-sampling
b) Up-sampling
c) Sampling
d) None of the mentioned
View Answer
Answer: a
Explanation: If the signal x(n) was originally obtained by sampling a signal xa(t), then x(n)=xa(nT). Now, y(n)=x(2n)(say)=xa(2nT). Hence the time scaling operation is equivalent to changing the sampling rate from 1/T to 1/2T, that is to decrease the rate by a factor of 2. So, time scaling is also called as down-sampling.
9. What is the condition for a signal x(n)=Brn where r=eαT to be called as an decaying exponential signal?
a) 0<r<∞
b) 0<r<1
c) r>1
d) r<0
View Answer
Answer: b
Explanation: When the value of ‘r’ lies between 0 and 1 then the value of x(n) goes on decreasing exponentially with increase in value of ‘n’. So, the signal is called as decaying exponential signal.
10. The function given by the equation x(n)=1, for n=0;=0, for n≠0 is a:
a) Step function
b) Ramp function
c) Triangular function
d) Impulse function
View Answer
Answer: d
Explanation: According to the definition of the impulse function, it is defined only at n=0 and is not defined elsewhere which is as per the signal given

1. Resolve the sequence1ainto a sum of weighted impulse sequences.
a) 2δ(n)+4δ(n-1)+3δ(n-3)
b) 2δ(n+1)+4δ(n)+3δ(n-2)
c) 2δ(n)+4δ(n-1)+3δ(n-2)
d) None of the mentioned
View Answer
Answer: b
Explanation: We know that, x(n)δ(n-k)=x(k)δ(n-k)
x(-1)=2=2δ(n+1)
x(0)=4=4δ(n)
x(2)=3=3δ(n-2)
Therefore, x(n)= 2δ(n+1)+4δ(n)+3δ(n-2)
2. The formula2that gives the response y(n) of the LTI system as the function of the input signal x(n) and the unit sample response h(n) is known as:
a) Convolution sum
b) Convolution product
c) Convolution Difference
d) None of the mentioned
View Answer
Answer: a
Explanation: The input x(n) is convoluted with the impulse response h(n) to yield the output y(n).As we are summing the different values, we call it as Convolution sum.
3. What is the order of the four operations that are needed to be done on h(k) in order to convolute x(k) and h(k)?
Step-1:Folding
Step-2:Multiplicaton with x(k)
Step-3:Shifting
Step-4:Summation
a) 1-2-3-4
b) 1-2-4-3
c) 2-1-3-4
d) 1-3-2-4
View Answer
Answer: d
Explanation: First the signal h(k) is folded to get h(-k). Then it is shifted by n to get h(n-k). Then it is multiplied by x(k) and then summed over -∞ to ∞.
4. The impulse response of a LTI system is h(n)={1,1,1}. What is the response of the signal to the input x(n)={1,2,3}?
a) {1,3,6,3,1}
b) {1,2,3,2,1}
c) {1,3,6,5,3}
d) {1,1,1,0,0}
View Answer
Answer: c
Explanation: Let y(n)=x(n)*h(n)(‘*’ symbol indicates convolution symbol)
From the formula of convolution we get,
y(0)=x(0)h(0)=1.1=1
y(1)=x(0)h(1)+x(1)h(0)=1.1+2.1=3
y(2)=x(0)h(2)+x(1)h(1)+x(2)h(0)=1.1+2.1+3.1=6
y(3)=x(1)h(2)+x(2)h(1)=2.1+3.1=5
y(4)=x(2)h(2)=3.1=3
Therefore, y(n)=x(n)*h(n)={1,3,6,5,3}
5. Determine the output y(n) of a LTI system with impulse response h(n)=anu(n),|a|<1with the input sequence x(n)=u(n).
a) (1-a(n+1))/(1-a)
b) (1-a(n-1))/(1-a)
c) (1+a(n+1))/(1+a)
d) None of the mentioned
View Answer
Answer: a
Explanation: Now fold the signal x(n) and shift it by one unit at a time and sum as follows
y(0)=x(0)h(0)=1
y(1)=h(0)x(1)+h(1)x(0)=1.1+a.1=1+a
y(2)=h(0)x(2)+h(1)x(1)+h(2)x(0)=1.1+a.1+a2.1=1+a+a2
Similarly, y(n)=1+a+a2+….an= (1-a(n+1))/(1-a)
6. x(n)*(h1(n)*h2(n))=(x(n)*h1(n))*h2(n)
a) True
b) False
View Answer
Answer: a
Explanation: According to the properties of convolution, Convolution of three signals obeys Associative property.
7. Determine the impulse response for the cascade of two LTI systems having impulse responses h1(n)=(1/2)2 u(n) and h2(n)= (1/4)2 u(n).
a) (1/2)n[2-(1/2)n], n<0
b) (1/2)n[2-(1/2)n], n>0
c) (1/2)n[2+(1/2)n], n<0
d) (1/2)n[2+(1/2)n], n>0
View Answer
Answer: b
Explanation: Let h2(n) be shifted and folded.
so, h(k)=h1(n)*h2(n)=7c
For k<0, h1(n)= h2(n)=0 since the unit step function is defined only on the right hand side.7d
8. x(n)*[h1(n)+h2(n)]=x(n)*h1(n)+x(n)*h2(n)
a) True
b) False
View Answer
Answer: a
Explanation: According to the properties of the convolution, convolution exhibits distributive property.
9. An LTI system is said to be causal if and only if
a) Impulse response is non-zero for positive values of n
b) Impulse response is zero for positive values of n
c) Impulse response is non-zero for negative values of n
d) Impulse response is zero for negative values of n
View Answer
Answer: d
Explanation: Let us consider a LTI system having an output at time n=n0given by the convolution formula
9
=(h(0)x(n0)+h(1)x(n0-1)+h(2)x(n0-2)+….)+(h(-1)x(n0+1)+h(-2)x(n0+2)+…)
As per the definition of the causality, the output should depend only on the present and past values of the input. So, the coefficients of the terms x(n0+1), x(n0+2)…. should be equal to zero.
that is, h(n)=0 for n<0 .
10. x(n)*δ(n-n0)=
a) x(n+n0)
b) x(n-n0)
c) x(-n-n0)
d) x(-n+n0)
View Answer
Answer: b
Explanation:
10
11. Is the system with impulse response h(n)=2nu(n-1) stable?
a) True
b) False
View Answer
Answer: b
Explanation:
11
So, the system is not stable.


1. What is the unit step response of the system described by the difference equation
y(n)=0.9y(n-1)-0.81y(n-2)+x(n) under the initial conditions y(-1)=y(-2)=0?
a) [1.099+1.088(0.9)n.cos(Ï€n/3+5.2o)]u(n)
b) [1.099+1.088(0.9)n.cos(Ï€n/3-5.2o)]u(n)
c) [1.099+1.088(0.9)n.cos(Ï€n/3-5.2o)].
d) None of the mentioned
View Answer
Answer: b
Explanation: The system function is H(z)=1/(1-0.9z-1+0.81z-2 )
The system has two complex-conjugate poles at p1=0.9ejÏ€/3 and p2=0.9e -jÏ€/3
The z-transform of the unit step sequence is
X(z)=1/(1-z-1 )
Therefore,
Yzs(z) = 1/((1-0.9e^(jÏ€/3) z-1)(1-0.9e-jÏ€/3 z-1 )(1-z-1))
= (0.542-j0.049)/((1-0.9ejÏ€/3 z-1) ) + (0.542-j0.049)/((1-0.9e^(jÏ€/3) z-1 ) ) + 1.099/(1-z-1 )
and hence the zero state response is yzs(n)= [1.099+1.088(0.9)n.cos(Ï€n/3-5.2o)]u(n)
Since the initial conditions are zero in this case, we can conclude that y(n)= yzs(n).
2. If all the poles of H(z) are outside the unit circle, then the system is said to be:
a) Only causal
b) Only BIBO stable
c) BIBO stable and causal
d) None of the mentioned
View Answer
Answer: d
Explanation: If all the poles of H(z) are outside an unit circle, it means that the system is neither causal nor BIBO stable.
3. If pk, k=1,2,…N are the poles of the system and |pk| < 1 for all k, then the natural response of such a system is called as Transient response.
a) True
b) False
View Answer
Answer: a
Explanation: If |pk| < 1 for all k, then ynr(n) decays to 0 as n approaches infinity. In such a case we refer to the natural response of the system as the transient response.
4. If all the poles have small magnitudes, then the rate of decay of signal is:
a) Slow
b) Constant
c) Rapid
d) None of the mentioned
View Answer
Answer: c
Explanation: If the magnitudes of the poles of the response of any system is very small i.e., almost equal to zero, then the system decays very rapidly.
5. If one or more poles are located near the unit circle , then the rate of decay of signal is:
a) Slow
b) Constant
c) Rapid
d) None of the mentioned
View Answer
Answer: a
Explanation: If the magnitudes of the poles of the response of any system is almost equal to one, then the system decays very slowly or the transient will persist for a relatively long time.
6. What is the transient response of the system described by the difference equation y(n)=0.5y(n-1)+x(n) when the input signal is x(n)= 10cos(Ï€n/4)u(n) and the system is initially at rest?
a) (0.5)nu(n)
b) 0.5(6.3)nu(n)
c) 6.3(0.5)n
d) 6.3(0.5)nu(n)
View Answer
Answer: d
Explanation: The system function for the system is
6aand therefore the system has a pole at z=0.5. The z-transform of the input signal is
6b
The natural or transient response is
ynr(n)= 6.3(0.5)nu(n)
7. What is the steady-state response of the system described by the difference equation y(n)=0.5y(n-1)+x(n) when the input signal is x(n)= 10cos(Ï€n/4)u(n) and the system is initially at rest?
a) 13.56cos(Ï€n/4 -28.7o)
b) 13.56cos(Ï€n/4 +28.7o)u(n)
c) 13.56cos(Ï€n/4 -28.7o)u(n)
d) None of the mentioned
View Answer
Answer: c
Explanation: The system function for the system is
7b
8. If the ROC of the system function is the exterior of a circle of radius r < ∞, including the point z = ∞, then the system is said to be:
a) Stable
b) Causal
c) Anti causal
d) None of the mentioned
View Answer
Answer: b
Explanation: A linear time invariant system is said to be causal if and only if the ROC of the system function is the exterior of a circle of radius r < ∞, including the point z = ∞.
9. A linear time invariant system is said to be BIBO stable if and only if the ROC of the system function:
a) Includes unit circle
b) Excludes unit circle
c) Is an unit circle
d) None of the mentioned
View Answer
Answer: a
Explanation: For an LTI system, if the ROC of the system function includes the unit circle, then the systm is said to be BIBO stable.
10. If all the poles of H(z) are inside the unit circle, then the system is said to be:
a) Only causal
b) Only BIBO stable
c) BIBO stable and causal
d) None of the mentioned
View Answer
Answer: c
Explanation: If all the poles of H(z) are inside an unit circle, then it follows the condition that |z|>r < 1, it means that the system is both causal and BIBO stable.
11. A linear time invariant system is characterized by the system function H(z)=1/(1-0.5z-1)+2/(1-3z-1 ).What is the h(n) if the system is stable?
a) (0.5)nu(n)-2(3)nu(n)
b) (0.5)nu(-n-1)-2(3)nu(-n-1)
c) (0.5)nu(-n-1)-2(3)nu(n)
d) (0.5)nu(n)-2(3)nu(-n-1)
View Answer
Answer: d
Explanation: The system has poles at z=0.5 and at z=3.
Since the system is stable, its ROC must include unit circle and hence it is 0.5<|z|<3 . Consequently, h(n) is non causal and is given as h(n)= (0.5)nu(n)-2(3)nu(-n-1)
12. A linear time invariant system is characterized by the system function H(z)=1/(1-0.5z-1)+2/(1-3z-1 ).What is the ROC of H(z) if the system is causal?
a) |z|<3 b) |z|>3
c) |z|<0.5 d) |z|>0.5
View Answer
Answer: b
Explanation: The system has poles at z=0.5 and at z=3.
Since the system is causal, its ROC is |z|>0.5 and |z|>3. The common region is |z|>3. So, ROC of given H(z) is |z|>3.
13. A linear time invariant system is characterized by the system function H(z)=1/(1-0.5z-1)+2/(1-3z-1 ).What is the h(n) if the system is anti causal?
a) (0.5)nu(n)+2(3)nu(n)
b) (0.5)nu(-n-1)-2(3)nu(-n-1)
c) -[(0.5)n+2(3)n]u(-n-1)
d) (0.5)nu(n)-2(3)nu(-n-1)
View Answer
Answer: c
Explanation: The system has poles at z=0.5 and at z=3.
If the system is anti causal, then the ROC is |z|<0.5.Hence h(n)= -[(0.5)n+2(3)n]u(-n-1)

1. What is the signal x(n) whose z-transform X(z)=log(1+az-1);|z|>|a|?
1a
View Answer
Answer: c
Explanation:
1b
2. If Z{x1(n)}=X1(z) and Z{x2(n)}=X2(z) then Z{x1(n)*x2(n)}=?
a) X1(z).X2(z)
b) X1(z)+X2(z)
c) X1(z)*X2(z)
d) None of the mentioned
View Answer
Answer: a
Explanation: According to the convolution property of z-transform, the z-transform of convolution of two sequences is the product of their respective z-transforms.
3. What is the convolution x(n) of the signals x1(n)={1,-2,1} and x2(n)={1,1,1,1,1,1}?
a) {1,1,0,0,0,0,1,1}
b) {-1,-1,0,0,0,0,-1,-1}
c) {-1,1,0,0,0,0,1,-1}
d) {1,-1,0,0,0,0,-1,1}
View Answer
Answer: d
Explanation:
3a
4. If Z{x1(n)}=X1(z) and Z{x2(n)}=X2(z) then what is the z-transform of correlation between the two signals?
a) X1(z).X2(z-1)
b) X1(z).X2(z-1)
c) X1(z).X2(z)
d) X1(z).X2(-z)
View Answer
Answer: b
Explanation: We know that rx1x2(l)=x1(l)*x2(-l)
Now Rx1x2(z)=Z{x1(l)}.Z{x2(-l)}=X1(z).X2(z-1)
5. If x(n) is causal, then lim┬(z→∞)X(z)=?
a) x(-1)
b) x(1)
c) x(0)
d) Cannot be determined
View Answer
Answer: c
Explanation: According to the initial value theorem, X(z)=x(0)+x(1)z -1+x(2)z-2+….
When z→∞, z -n tends to 0 because n>0.
So lim┬(n→∞)⁡〖X(z)〗=x(0).
6. If Z{x(n)}=X(z) and the poles of X(z) are all inside the unit circle, then the final value of x(n) as n→∞ is given by i.e., lim┬(n→∞)x(n)=?
6b
View Answer
Answer: a
Explanation: According to the Final Value theorem of z-transform we have,
6c
7. What is the z-transform of the signal x(n)=δ(n-n0)?
a) zn0
b) z-n0
c) zn-n0
d) zn+n0
View Answer
Answer: b
Explanation: From the definition of z-transform,
7
8. If X(z) is the z-transform of the signal x(n), then what is the z-transform of x*(n)?
a) X(z*)
b) X*(z)
c) X*(-z)
d) X*(z*)
View Answer
Answer: d
Explanation: According to the conjugation property of z-transform, we have
Z{x*(n)}= X*(z*)
9. If x(n) is an imaginary sequence, then the z-transform of the real part of the sequence is:
a) 1/2[X(z)+X*(z*)] b) 1/2[X(z)-X*(z*)] c) 1/2[X(-z)-X*(z*)] d) 1/2[X(-z)+X*(z*)] View Answer
Answer: a
Explanation: If x(N) is an imaginary sequence, then the real part of x(n) is given as
Real{x(n)}= 1/2[x(n)+x*(n)] According to linearity property of z-transform, we get
Z{ Real{x(n)}}= 1/2[X(z)+X*(z*)]
10. What is the signal whose z-transform is given as10a
a) x1(n)*x2(n)
b) x1(n)*x2(-n)
c) x1(n).x2(n)
d) x1(n)*x2*(n)
View Answer
Answer: c
Explanation: From the convolution property in z-domain we have,
10b
11. What is the z-transform of the signal x(n)= x1(n).x2*(n)?
11a
View Answer
Answer: b
Explanation:
We know that Z{x*(n)}=X*(z*)
Now from the multiplication property in time domain we get,
10b
12. If x1(n)={1,2,3} and x2(n)={1,1,1}, then what is the convolution sequence of the given two signals?
a) {1,2,3,1,1}
b) {1,2,3,4,5}
c) {1,3,5,6,2}
d) {1,2,6,5,3}
View Answer
Answer: d
Explanation: Given x1(n)={ 1,2,3}=>X1(z)=1+2z -1+3z -2
x2(n){1,1,1}=>X2(z)=1+z -1+z -2
Now from the convolution in time domain property of z-transform, we have
Z{ x1(n)* x2(n)}= X1(z). X2(z)
=> X(z)=1+2z-1+6z-2+5z-3+3z-4 =>x(n)={1,2,6,5,3}
13. What is the z-transform of the signal x(n)=cos(jω0n)u(n)?
13a
View Answer
Answer: c
Explanation: By Euler’s identity, the given signal x(n) can be written as
13b
14. What is the z-transform of the signal defined as x(n)=u(n)-u(n+N)?
14a
View Answer
Answer: d
Explanation:
14b
15. What is the z-transform of the signal x(n)=[5(3n)-9(7n)]u(n)?
15a
View Answer
Answer: a
Explanation:
15b
1. Which of the following method is used to find the inverse z-transform of a signal?
a) Counter integration
b) Expansion into a series of terms
c) Partial fraction expansion
d) All of the mentioned
View Answer
Answer: d
Explanation: All the methods mentioned above can be used to calculate the inverse z-transform of the given signal.
2. What is the inverse z-transform of X(z)=1/(1-1.5z-1+0.5z-2 ) if ROC is |z|>1?
a) {1,3/2,7/4,15/8,31/16,….}
a) {1,2/3,4/7,8/15,16/31,….}
a) {1/2,3/4,7/8,15/16,31/32,….}
d) None of the mentioned
View Answer
Answer: a
Explanation: Since the ROC is the exterior circle, we expect x(n) to be a causal signal. Thus we seek a power series expansion in negative powers of ‘z’. By dividing the numerator of X(z) by its denominator, we obtain the power series
2b
So, we obtain x(n)= {1,3/2,7/4,15/8,31/16,….}.
3. What is the inverse z-transform of X(z)=1/(1-1.5z-1+0.5z-2 ) if ROC is |z| < 0.5?
a) {….62,30,14,6,2}
b) {…..62,30,14,6,2,0,0}
c) {0,0,2,6,14,30,62…..}
d) {2,6,14,30,62…..}
View Answer
Answer: b
Explanation: In this case the ROC is the interior of a circle. Consequently, the signal x(n) is anti causal. To obtain a power series expansion in positive powers of z, we perform the long division in the following way:
3a
Thus3b
In this case x(n)=0 for n≥0.Thus we obtain x(n)= {…..62,30,14,6,2,0,0}
4. What is the inverse z-transform of X(z)=log(1+az-1) |z|>|a|?
4ad)None of the mentioned
View Answer
Answer: c
Explanation: Using the power series expansion for log(1+x), with |x|<1, we have4b
5. What is the proper fraction and polynomial form of the improper rational transform
X(z)= (1+3z-1+11/6 z-2+1/3 z-3)/(1+5/6 z-1+1/6 z-2 )?
a) 1+2z -1+(1/6 z-1)/(1+5/6 z-1+1/6 z-2 )
b) 1-2z -1+(1/6 z-1)/(1+5/6 z-1+1/6 z-2 )
c) 1+2z -1+(1/3 z-1)/(1+5/6 z-1+1/6 z-2)
d) 1+2z -1-(1/6 z-1)/(1+5/6 z-1+1/6 z-2 )
View Answer
Answer: a
Explanation: First, we note that we should reduce the numerator so that the terms z-2 and z -3 are eliminated. Thus we should carry out the long division with these two polynomials written in the reverse order. We stop the division when the order of the remainder becomes z -1. Then we obtain
X(z)= 1+2z -1+(1/6 z-1)/(1+5/6 z-1+1/6 z-2 )
6. What is the partial fraction expansion of the proper function X(z)= 1/(1-1.5z-1+0.5z-2 )?
a) 2z/(z-1)-z/(z+0.5)
b) 2z/(z-1)+z/(z-0.5)
c) 2z/(z-1)+z/(z+0.5)
d) 2z/(z-1)-z/(z-0.5)
View Answer
Answer: d
Explanation: First we eliminate the negative powers of z by multiplying both numerator and denominator by z2.
Thus we obtain X(z)= z2/(z2-1.5z+0.5)
The poles of X(z) are p1=1 and p2=0.5. Consequently, the expansion will be
(X(z))/z = z/((z-1)(z-0.5)) = 2/((z-1) ) – 1/((z-0.5) )( obtained by applying partial fractions)
=>X(z)= 2z/(z-1)-z/(z-0.5)
7. What is the partial fraction expansion of X(z)= (1+z-1)/(1-z-1+0.5z-2 )?
a) (z(0.5-1.5j))/(z-0.5-0.5j) – (z(0.5+1.5j))/(z-0.5+0.5j)
b) (z(0.5-1.5j))/(z-0.5-0.5j) + (z(0.5+1.5j))/(z-0.5+0.5j)
c) (z(0.5+1.5j))/(z-0.5-0.5j) – (z(0.5-1.5j))/(z-0.5+0.5j)
d) (z(0.5+1.5j))/(z-0.5-0.5j) + (z(0.5-1.5j))/(z-0.5+0.5j)
View Answer
Answer: b
Explanation: To eliminate the negative powers of z, we multiply both numerator and denominator by z2. Thus,
X(z)=(z(z+1))/(z-2-z+0.5)
The poles of X(z) are complex conjugates p1=0.5+0.5j and p2=0.5-0.5j
Consequently the expansion will be
X(z)= (z(0.5-1.5j))/(z-0.5-0.5j) + (z(0.5+1.5j))/(z-0.5+0.5j)
8. What is the partial fraction expansion of X(z)=1/((1+z-1 )(1-z-1)2)?
a) z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z+1)〗2 )
b) z/(4(z+1)) + 3z/(4(z-1)) – z/(2〖(z+1)〗2 )
c) z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z-1)〗2 )
d) z/(4(z+1)) + z/(4(z-1)) + z/(2〖(z+1)〗2 )
View Answer
Answer: c
Explanation: First we express X(z) in terms of positive powers of z, in the form X(z)=z3/((z+1)〖(z-1)〗2 )
X(z) has a simple pole at z=-1 and a double pole at z=1. In such a case the approximate partial fraction expansion is
(X(z))/z = z2/((z+1)〖(z-1)〗2 ) =A/(z+1) + B/(z-1) + C/〖(z-1)〗2
On simplifying, we get the values of A, B and C as 1/4, 3/4 and 1/2 respectively.
Therefore, we get X(z)= z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z-1)〗2 ) .
9. What is the inverse z-transform of X(z)= 1/(1-1.5z-1+0.5z2-2 ) if ROC is |z|>1?
a) (2-0.5n)u(n)
b) (2+0.5n)u(n)
c) (2n-0.5n)u(n)
d) None of the mentioned
View Answer
Answer: a
Explanation: The partial fraction expansion for the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In case when ROC is |z|>1, the signal x(n) is causal and both the terms in the above equation are causal terms. Thus, when we apply inverse z-transform to the above equation, we get
x(n)=2(1)nu(n)-(0.5)nu(n)=(2-0.5n)u(n)
10. What is the inverse z-transform of X(z)= 1/(1-1.5z-1+0.5z-2 ) if ROC is |z|<0.5? a) [-2-0.5n]u(n)
b) [-2+0.5n]u(n)
c) [-2+0.5n]u(-n-1)
d) [-2-0.5n]u(-n-1)
View Answer
Answer: c
Explanation: The partial fraction expansion for the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In case when ROC is |z|<0.5,the signal is anti causal. Thus both the terms in the above equation are anti causal terms. So, if we apply inverse z-transform to the above equation we get x(n)= [-2+0.5n]u(-n-1)
11. What is the inverse z-transform of X(z)= 1/(1-1.5z-1+0.5z-2 ) if ROC is 0.5<|z|<1? a) -2u(-n-1)+(0.5)nu(n)
b) -2u(-n-1)-(0.5)nu(n)
c) -2u(-n-1)+(0.5)nu(-n-1)
d) 2u(n)+(0.5)nu(-n-1)
View Answer
Answer: b
Explanation: The partial fraction expansion of the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In this case ROC is 0.5<|z|<1 is a ring, which implies that the signal is two sided. Thus one of the signal corresponds to a causal signal and the other corresponds to an anti causal signal. Obviously, the ROC given is the overlapping of the regions |z|>0.5 and |z|<1. Hence the pole p2=0.5 provides the causal part and the pole p1=1 provides the anti causal part. SO, if we apply the inverse z-transform we get x(n)= -2u(-n-1)-(0.5)nu(n)
12.What is the causal signal x(n) having the z-transform X(z)= 1/((1+z-1 ) [(1-z-1)]2 )?
a)[1/4(-1)n+3/4-n/2]u(n)
b)[1/4(-1)n+3/4-n/2]u(-n-1)
c)[1/4+3/4(-1)n-n/2]u(n)
d)[1/4(-1)n+3/4+n/2]u(n)
View Answer
Answer: d
Explanation: The partial fraction expansion of X(z) is X(z)= z/(4(z+1)) + 3z/(4(z-1)) + z/(2[(z-1)]2)
When we apply the inverse z-transform for the above equation, we get
x(n)=[1/4(-1)n+3/4+n/2]u(n)