The DTFT is obtained from the z-transform by evaluating it on the unit circle, i.e., , making the z-transform a generalization of the DTFT.
Incorrect! Try again.
5Evaluating the z-transform on the unit circle yields the:
the z-transform
Easy
A.DTFT of the signal
B.Laplace transform
C.Continuous Fourier transform
D.Autocorrelation of the signal
Correct Answer: DTFT of the signal
Explanation:
On the unit circle , the z-transform reduces to the Discrete-Time Fourier Transform of the signal.
Incorrect! Try again.
6The Region of Convergence (ROC) of a z-transform is the set of values of for which:
The region of convergence for the z transform
Easy
A.The signal is causal
B.The summation converges
C.The system is unstable
D.The poles lie on the unit circle
Correct Answer: The summation converges
Explanation:
The ROC is defined as the region in the complex z-plane where the z-transform summation converges to a finite value.
Incorrect! Try again.
7In the z-plane, the ROC is typically bounded by:
The region of convergence for the z transform
Easy
A.Straight horizontal lines
B.Parabolas through the origin
C.Circles centered at the origin
D.Straight vertical lines
Correct Answer: Circles centered at the origin
Explanation:
Since convergence depends on the magnitude , the ROC is bounded by circles (constant ) centered at the origin of the z-plane.
Incorrect! Try again.
8The ROC of a z-transform can contain:
The region of convergence for the z transform
Easy
A.No poles
B.Exactly one pole
C.All the poles
D.Only the zeros
Correct Answer: No poles
Explanation:
By definition, the z-transform does not converge at its poles, so the ROC never includes any pole of .
Incorrect! Try again.
9For a right-sided (causal) sequence, the ROC is:
The region of convergence for the z transform
Easy
A.The interior of a circle
B.The entire z-plane
C.A ring between two circles
D.The exterior of a circle
Correct Answer: The exterior of a circle
Explanation:
A right-sided sequence has an ROC that is the exterior of a circle, i.e., , extending outward from the outermost pole.
Incorrect! Try again.
10For a left-sided sequence, the ROC is:
The region of convergence for the z transform
Easy
A.A ring between two circles
B.The exterior of a circle
C.The unit circle only
D.The interior of a circle
Correct Answer: The interior of a circle
Explanation:
A left-sided sequence has an ROC that is the interior of a circle, i.e., , extending inward from the innermost pole.
Incorrect! Try again.
11The z-transform is a __ operation with respect to its input signals.
Properties of the z-transform
Easy
A.Logarithmic
B.Linear
C.Quadratic
D.Nonlinear
Correct Answer: Linear
Explanation:
The z-transform satisfies linearity: .
Incorrect! Try again.
12A time shift of corresponds to multiplication of by:
Properties of the z-transform
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The time-shifting property states that .
Incorrect! Try again.
13Convolution of two signals in the time domain corresponds to which operation in the z-domain?
Properties of the z-transform
Easy
A.Division
B.Convolution
C.Addition
D.Multiplication
Correct Answer: Multiplication
Explanation:
The convolution property states that , i.e., convolution becomes multiplication in the z-domain.
Incorrect! Try again.
14The property is known as:
Properties of the z-transform
Easy
A.Time reversal
B.Differentiation
C.Conjugation
D.Scaling in the z-domain
Correct Answer: Scaling in the z-domain
Explanation:
Multiplication by in time corresponds to scaling in the z-domain: .
Incorrect! Try again.
15A common method to compute the inverse z-transform of a rational function is:
The inverse z transform
Easy
A.Gaussian elimination
B.Partial fraction expansion
C.Numerical integration only
D.Fourier series expansion
Correct Answer: Partial fraction expansion
Explanation:
Partial fraction expansion breaks a rational into simpler terms whose inverse z-transforms are known from standard tables.
Incorrect! Try again.
16Which of the following is NOT a standard method for finding the inverse z-transform?
The inverse z transform
Easy
A.Partial fraction expansion
B.Contour integration
C.Power series (long division)
D.Laplace inversion
Correct Answer: Laplace inversion
Explanation:
Standard inverse z-transform methods include partial fractions, power series expansion, and contour integration. Laplace inversion applies to the Laplace transform, not the z-transform.
Incorrect! Try again.
17To uniquely determine the inverse z-transform from , one must also know the:
The inverse z transform
Easy
A.Phase spectrum
B.Signal power
C.ROC
D.Sampling rate
Correct Answer: ROC
Explanation:
Different signals can share the same expression, so the ROC is required to uniquely determine the corresponding time-domain sequence.
Incorrect! Try again.
18For a discrete-time LTI system, the z-transform of the impulse response is called the:
Analysis and characterisation of LTI systems using z-transforms
Easy
A.Impulse spectrum
B.Step response
C.Transfer function
D.Frequency deviation
Correct Answer: Transfer function
Explanation:
The z-transform of the impulse response gives the system transfer function , which characterizes the LTI system.
Incorrect! Try again.
19A causal and stable discrete-time LTI system has all its poles located:
Analysis and characterisation of LTI systems using z-transforms
Easy
A.Inside the unit circle
B.At the origin only
C.Outside the unit circle
D.On the unit circle
Correct Answer: Inside the unit circle
Explanation:
For a causal LTI system to be stable, all poles of must lie strictly inside the unit circle so that the ROC includes the unit circle.
Incorrect! Try again.
20In a pole-zero plot of the z-plane, zeros are usually marked with the symbol:
Software simulation of system representation and pole zero analysis
Easy
A.A cross ()
B.A square
C.A triangle
D.A circle ()
Correct Answer: A circle ()
Explanation:
By convention, zeros are represented by circles () and poles are represented by crosses () in a pole-zero plot.
Incorrect! Try again.
21The z-transform of the discrete-time signal is:
the z-transform
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
For , , valid for .
Incorrect! Try again.
22The z-transform of the unit impulse is:
the z-transform
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Since is nonzero only at , , with ROC being the entire z-plane.
Incorrect! Try again.
23For a right-sided sequence, the region of convergence (ROC) of the z-transform is:
The region of convergence for the z transform
Medium
A.The entire z-plane except infinity
B.An annular ring:
C.The interior of a circle:
D.The exterior of a circle:
Correct Answer: The exterior of a circle:
Explanation:
A right-sided (causal-type) sequence has an ROC that extends outward from the outermost pole, i.e., .
Incorrect! Try again.
24The z-transform corresponds to a causal signal. Its ROC is:
The region of convergence for the z transform
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The pole is at . For a causal (right-sided) sequence, the ROC lies outside the outermost pole, so .
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25Which condition on the ROC guarantees that an LTI system is stable?
The region of convergence for the z transform
Medium
A.The ROC extends to
B.The ROC includes the origin
C.The ROC includes the unit circle
D.The ROC excludes all poles and the unit circle
Correct Answer: The ROC includes the unit circle
Explanation:
A discrete-time LTI system is BIBO stable if and only if the ROC of its transfer function includes the unit circle.
Incorrect! Try again.
26If , then the z-transform of is:
Properties of the z-transform
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The time-shift property states that a delay by multiplies the z-transform by .
Incorrect! Try again.
27Using the convolution property, if , then equals:
Properties of the z-transform
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Convolution in the time domain becomes multiplication in the z-domain: .
Incorrect! Try again.
28If , the z-transform of (scaling in the z-domain) is:
Properties of the z-transform
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Multiplication by in time scales the argument: .
Incorrect! Try again.
29The z-transform differentiation property gives the transform of as:
Properties of the z-transform
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The differentiation-in-z property states .
Incorrect! Try again.
30The inverse z-transform of , ROC , is:
The inverse z transform
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The pole is at and the ROC is outside it, giving the causal (right-sided) sequence .
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31For with ROC , the inverse z-transform is:
The inverse z transform
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Since the ROC is inside the pole, the sequence is left-sided: .
Incorrect! Try again.
32Which technique is most commonly used to invert a rational z-transform with distinct poles?
The inverse z transform
Medium
A.Partial fraction expansion
B.Laplace transform inversion
C.Fourier series expansion
D.Gaussian elimination
Correct Answer: Partial fraction expansion
Explanation:
Rational is decomposed into simple terms via partial fraction expansion, each inverted using known transform pairs and the ROC.
Incorrect! Try again.
33Using long division (power series method) on yields a sequence corresponding to:
The inverse z transform
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Long division gives , whose coefficients are all 1 for , i.e., the unit step .
Incorrect! Try again.
34An LTI system has transfer function . The system is causal and:
Analysis and characterisation of LTI systems using z-transforms
Medium
A.Unstable, since the pole lies outside the unit circle
B.Marginally stable, since the pole lies on the unit circle
C.Stable, since the pole lies inside the unit circle
D.Stable only for negative inputs
Correct Answer: Stable, since the pole lies inside the unit circle
Explanation:
The pole at has magnitude less than 1, so for a causal system the ROC includes the unit circle, making it stable.
Incorrect! Try again.
35A causal LTI system is described by . Its transfer function is:
Analysis and characterisation of LTI systems using z-transforms
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Taking the z-transform: , so .
Incorrect! Try again.
36For a causal LTI system with , the poles are located at:
Analysis and characterisation of LTI systems using z-transforms
Medium
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
Setting gives roots and . A pole on the unit circle means the system is not stable.
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37The frequency response of a stable LTI system is obtained from by evaluating it at:
Analysis and characterisation of LTI systems using z-transforms
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The DTFT (frequency response) equals evaluated on the unit circle, , provided the ROC includes it.
Incorrect! Try again.
38In MATLAB, the command zplane(b, a) where b and a are numerator and denominator coefficients is used to:
Software simulation of system representation and pole zero analysis
Medium
A.Plot the magnitude frequency response only
B.Plot the poles and zeros of a system in the z-plane
C.Perform the inverse z-transform numerically
D.Compute the impulse response of the system
Correct Answer: Plot the poles and zeros of a system in the z-plane
Explanation:
The zplane function displays the pole-zero plot along with the unit circle for a system defined by coefficients b and a.
Incorrect! Try again.
39In a pole-zero plot, poles are conventionally marked with and zeros with . If all poles lie inside the unit circle for a causal system, the system is:
Software simulation of system representation and pole zero analysis
Medium
A.Marginally stable
B.Stable
C.Non-causal
D.Unstable
Correct Answer: Stable
Explanation:
For a causal system, having all poles strictly inside the unit circle ensures the ROC includes the unit circle, guaranteeing stability.
Incorrect! Try again.
40Given transfer function coefficients, which MATLAB function converts them into pole-zero-gain form?
Software simulation of system representation and pole zero analysis
Medium
A.freqz
B.conv
C.filter
D.tf2zp
Correct Answer: tf2zp
Explanation:
The tf2zp function converts transfer-function (numerator/denominator) representation into zeros, poles, and gain form for analysis.
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41Consider the two-sided signal with . What is its bilateral z-transform and ROC?
the z-transform
Hard
A., ROC:
B., ROC:
C., ROC:
D., ROC:
Correct Answer: , ROC:
Explanation:
Split as . The first term gives (), the second gives (). Combining yields with a ring ROC .
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42A signal has z-transform with poles at and . Which ROC corresponds to a stable but non-causal system?
The region of convergence for the z transform
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Stability requires the ROC to include the unit circle . Only the annulus contains . Since it is a two-sided ROC (not outermost pole), the system is non-causal but stable.
Incorrect! Try again.
43Given with ROC , find .
The inverse z transform
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Partial fractions: . So . With (causal), inverse gives .
Incorrect! Try again.
44If , what is the z-transform of ?
Properties of the z-transform
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The differentiation property gives . Applying it twice: maps to .
Incorrect! Try again.
45An LTI system has . Which statement about stability and minimum-phase is correct?
Analysis and characterisation of LTI systems using z-transforms
Hard
A.The system is unstable because it has a zero at
B.The system is both stable and minimum-phase with all-pass characteristics
C.The system is stable and causal but not minimum-phase (zero outside unit circle)
D.The system is minimum-phase since the pole is inside the unit circle
Correct Answer: The system is stable and causal but not minimum-phase (zero outside unit circle)
Explanation:
The pole at is inside the unit circle, so a causal system is stable. Minimum-phase requires all zeros inside the unit circle; here the zero is at (outside), so the system is not minimum-phase.
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46Using the initial value theorem, find for the causal signal with .
Properties of the z-transform
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The initial value theorem states . Dividing numerator and denominator by : .
Incorrect! Try again.
47For with , the inverse z-transform for is:
The inverse z transform
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Using the series with : , so for .
Incorrect! Try again.
48A causal LTI system is described by . What are the pole locations?
Analysis and characterisation of LTI systems using z-transforms
Hard
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
The characteristic equation is . Factoring: , giving poles at and , both inside the unit circle (stable causal system).
Incorrect! Try again.
49If with ROC and with ROC , the ROC of the convolution is at least:
Properties of the z-transform
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Convolution multiplies transforms, and the ROC of the product contains the intersection of individual ROCs. and intersect at . (Pole-zero cancellation could enlarge it, hence 'at least'.)
Incorrect! Try again.
50A finite-length signal is nonzero only for . What is the ROC of ?
The region of convergence for the z transform
Hard
A.Entire -plane including and
B. excluding the unit circle
C.Entire -plane except possibly and
D. only
Correct Answer: Entire -plane except possibly and
Explanation:
For a finite-duration signal, is a polynomial in and . Positive powers of (from ) make problematic; negative powers (from ) make problematic. The ROC is the whole plane except possibly these two points.
Incorrect! Try again.
51Find the inverse z-transform of , .
The inverse z transform
Hard
A.
B. for even , zero for odd
C. for all
D.
Correct Answer: for even , zero for odd
Explanation:
. Only even indices are nonzero with value ; odd samples are zero.
Incorrect! Try again.
52A system function is . How many poles and zeros are in the finite -plane (excluding origin/infinity trivial ones counted properly)?
Analysis and characterisation of LTI systems using z-transforms
Hard
A.poles at and ; zeros at and
B.2 poles at ; zeros at and
C.2 poles at ; one zero at
D.3 poles; 2 zeros
Correct Answer: 2 poles at ; zeros at and
Explanation:
Writing in positive powers, the denominator gives poles at and . The numerator contributes a zero at and, from the factor, effectively a zero at in the finite plane.
Incorrect! Try again.
53Applying the final value theorem, find for , ROC .
Properties of the z-transform
Hard
A.
B.The theorem does not apply here
C.
D.
Correct Answer:
Explanation:
The final value theorem applies since poles (other than at ) are inside the unit circle. ; at this equals .
Incorrect! Try again.
54The z-transform of is:
the z-transform
Hard
A.,
B.,
C.,
D.,
Correct Answer: ,
Explanation:
Start with . Apply differentiation: , ROC .
Incorrect! Try again.
55In MATLAB, given b = [1 -0.5] and a = [1 -1.5 0.56], which command correctly returns poles, zeros, and gain of the transfer function ?
Software simulation of system representation and pole zero analysis
Hard
A.[z,p] = roots(b,a)
B.[z,p,k] = residuez(b,a)
C.[z,p,k] = tf2zp(b,a)
D.[z,p,k] = zp2tf(b,a)
Correct Answer: [z,p,k] = tf2zp(b,a)
Explanation:
tf2zp converts a transfer-function representation (numerator b, denominator a) to zero-pole-gain form. zp2tf does the reverse, residuez gives partial-fraction residues, and roots operates on a single polynomial only.
Incorrect! Try again.
56A pole-zero plot generated by zplane shows poles at and a zero at . What can you infer about the system's magnitude response?
Software simulation of system representation and pole zero analysis
Hard
A.A resonant peak near normalized frequency due to poles near the unit circle
B.A flat all-pass response since gain is unity
C.A notch (null) at because the zero dominates
D.Instability because complex poles always cause growth
Correct Answer: A resonant peak near normalized frequency due to poles near the unit circle
Explanation:
Poles at radius (close to the unit circle) at angle produce a resonant peak in near . The system is stable since ; the zero at origin only adds linear phase/delay.
Incorrect! Try again.
57An all-pass system has . For this to be stable and all-pass, which condition on holds, and what is ?
Analysis and characterisation of LTI systems using z-transforms
Hard
A. real and
B. and
C. and for all
D. and
Correct Answer: and for all
Explanation:
Stability of a causal system requires the pole inside the unit circle: . For an all-pass section, the pole and zero are reciprocal conjugates, so the magnitude response is exactly unity at every frequency.
Incorrect! Try again.
58For with ROC , the inverse z-transform is:
The inverse z transform
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Partial fractions: . The annular ROC means the pole gives a causal term () and the pole gives an anti-causal term (), yielding the stated result.
Incorrect! Try again.
59Which statement best captures why the z-transform generalizes the discrete-time Fourier transform (DTFT)?
Introduction
Hard
A.The z-transform evaluates the signal on , and reduces to the DTFT on the unit circle when the ROC includes it
B.The DTFT is the z-transform evaluated at
C.The z-transform only exists for signals whose DTFT diverges everywhere
D.The z-transform is identical to the DTFT but uses instead of
Correct Answer: The z-transform evaluates the signal on , and reduces to the DTFT on the unit circle when the ROC includes it
Explanation:
Writing , the z-transform is the DTFT of . When (unit circle) lies in the ROC, the z-transform equals the DTFT, so the z-transform can converge for signals whose DTFT does not.
Incorrect! Try again.
60You simulate a filter with poles very close to the unit circle (radius ). What numerical/practical issue is most likely to appear in the impulse response computed via filter in floating point?
Software simulation of system representation and pole zero analysis
Hard
A.The impulse response becomes exactly periodic with no decay
B.The output is identically zero due to pole-zero cancellation
C.Very slowly decaying, long impulse response sensitive to coefficient quantization
D.The filter automatically becomes an ideal integrator
Correct Answer: Very slowly decaying, long impulse response sensitive to coefficient quantization
Explanation:
Poles at radius are nearly marginally stable, so the impulse response decays extremely slowly and requires many samples. Such near-unit-circle poles are highly sensitive to coefficient quantization, which can push them outside the unit circle and cause instability.
Incorrect! Try again.
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