Nepal Engineering Council · Electronics & Communication Engineering · Chapter 8
Signal System and Digital Signal Processing
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181 questions in 6 syllabus topics.
8.1 Signal and system
31 questions · AExE0801
1. A signal is best defined as:
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Think of what a microphone voltage or an image represents.
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Answer: A. A function of one or more independent variables that conveys information about a physical phenomenon
A signal is a function of independent variables (time, space, etc.) carrying information; a system is what processes it.
2. What is the total energy of the signal x(t) = e^(−2t)·u(t)?
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Square the signal before integrating.
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Answer: B. 0.25 J
E = ∫₀^∞ e^(−4t) dt = 1/4 = 0.25 J.
3. What is the average power of x(t) = 5 cos(100πt + π/3)?
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Average power of a sinusoid depends only on its amplitude.
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Answer: C. 12.5 W
For A cos(ωt + φ), the average power is A²/2 = 25/2 = 12.5 W; the phase does not matter.
4. What is the fundamental period of the discrete-time signal x[n] = cos(3πn/8)?
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N must be an integer, so find the smallest k that makes 16k/3 whole.
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Answer: D. 16 samples
N = 2πk/ω₀ = 16k/3; the smallest integer N occurs for k = 3, giving N = 16.
5. The discrete-time signal x[n] = cos(2n) is:
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A period in discrete time must be an integer number of samples.
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Answer: A. Not periodic, because ω₀/2π = 1/π is irrational
A discrete sinusoid is periodic only if ω₀/2π is a rational number; 2/(2π) = 1/π is irrational.
6. What is the value of ∫ (t² + 3)·δ(t − 2) dt taken over all time?
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The impulse picks out the value of the other factor at one instant.
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Answer: C. 7
By the sifting property the integral equals (t² + 3) at t = 2, i.e. 4 + 3 = 7.
7. The scaled impulse δ(2t) is equal to:
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Compare the areas under δ(2t) and δ(t).
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Answer: D. 0.5 δ(t)
δ(at) = δ(t)/|a|, so δ(2t) = δ(t)/2.
8. The unit step u(t) is related to the unit impulse δ(t) by:
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Think about the area accumulated by an impulse as time passes t = 0.
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Answer: B. u(t) is the running integral of δ(t)
u(t) = ∫ from −∞ to t of δ(τ) dτ, equivalently δ(t) = du(t)/dt.
9. The normalised sinc function sinc(x) = sin(πx)/(πx) has value 1 at x = 0 and is zero at:
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Where does sin(πx) vanish?
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Answer: D. All non-zero integer values of x
sin(πx) = 0 when x is an integer; at x = 0 the limit is 1, so the zeros are at x = ±1, ±2, ...
10. The signum function sgn(t) can be written in terms of the unit step as:
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Check the value of your expression for positive and for negative t.
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Answer: B. 2u(t) − 1
sgn(t) = +1 for t > 0 and −1 for t < 0, which equals 2u(t) − 1.
11. If x[n] = {1, 2, 1} (starting at n = 0) is applied to an LTI system with h[n] = {1, 1}, the output y[n] is:
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The output length is N₁ + N₂ − 1.
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Answer: C. {1, 3, 3, 1}
Linear convolution: y[0]=1, y[1]=2+1=3, y[2]=1+2=3, y[3]=1; the length is 3+2−1 = 4.
12. The convolution of the unit step with itself, u(t) * u(t), is:
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Convolving with u(t) is the same as integrating.
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Answer: C. t·u(t), the unit ramp
u(t)*u(t) = ∫₀^t 1 dτ = t for t ≥ 0, i.e. the unit ramp r(t) = t·u(t).
13. For an LTI system, the output for any input is completely determined by:
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Which single response lets you compute the output by convolution?
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Answer: B. Its impulse response
y(t) = x(t) * h(t); the impulse response fully characterises an LTI system.
14. Which of the following systems is linear and time-invariant?
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Test each one for additivity and for the effect of a shifted input.
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Answer: A. y(t) = 3 x(t − 2)
3x(t−2) is a scaled delay (LTI). x+2 fails homogeneity, t·x(t) and x(2t) are time-varying.
15. A real periodic signal has odd symmetry and half-wave symmetry (like a square wave centred on t = 0). Its trigonometric Fourier series contains:
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Apply the two symmetry rules one after the other.
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Answer: D. Only sine terms of odd harmonics
Odd symmetry removes a₀ and cosine terms; half-wave symmetry removes even harmonics, leaving odd-harmonic sines.
16. A periodic signal is x(t) = 3 + 4 cos(ω₀t) + 2 sin(3ω₀t). Its average power is:
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DC power is A², each sinusoid contributes A²/2.
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Answer: D. 19 W
By Parseval: P = 3² + 4²/2 + 2²/2 = 9 + 8 + 2 = 19 W.
17. Near a jump discontinuity, the partial sum of a Fourier series overshoots by about 9% of the jump no matter how many terms are taken. This is called:
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The effect is named after a physicist who explained it in 1899.
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Answer: B. Gibbs phenomenon
The persistent ≈9% overshoot near discontinuities is Gibbs phenomenon.
18. Which property holds for the exponential Fourier series coefficients cₖ of a real, even periodic signal?
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Combine conjugate symmetry with even symmetry.
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Answer: D. cₖ are real and even in k
A real signal gives c₋ₖ = cₖ*; even symmetry makes the coefficients real, so they are real and even.
19. The Fourier transform of x(t) = e^(−at)·u(t), a > 0, is:
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Combine the two exponents and integrate from 0 to ∞.
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Answer: C. 1/(a + jω)
X(ω) = ∫₀^∞ e^(−at) e^(−jωt) dt = 1/(a + jω).
20. The Fourier transform of the signum function sgn(t) is:
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The signum has zero average value, so no impulse at ω = 0.
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Answer: A. 2/(jω)
sgn(t) is the limit of e^(−a|t|)sgn(t); its transform tends to 2/(jω). Note u(t) = (1 + sgn t)/2 gives πδ(ω) + 1/(jω).
21. A rectangular pulse of width 2 ms is centred at t = 0. Its amplitude spectrum has the first zero crossing at:
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Nulls of the sinc occur at multiples of 1/T.
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Answer: A. 500 Hz
The transform is A·T·sinc(fT); the first null is at f = 1/T = 1/0.002 = 500 Hz.
22. A signal x(t) is band-limited to 4 kHz. The bandwidth of x(3t) is:
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Compression in time means expansion in frequency.
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Answer: B. 12 kHz
x(at) ↔ (1/|a|)X(ω/a); compressing time by 3 expands the spectrum by 3, giving 12 kHz.
23. Delaying a signal by t₀ seconds changes its Fourier transform X(ω) to:
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A delay should not change the magnitude spectrum.
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Answer: A. X(ω)·e^(−jωt₀)
The time-shift property: x(t − t₀) ↔ X(ω)e^(−jωt₀); only the phase changes.
24. The impulse response of an ideal Hilbert transformer is h(t) = 1/(πt). Its effect on the input spectrum is:
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Write the Hilbert transformer's frequency response as −j times something.
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Answer: D. A −90° phase shift for positive frequencies and +90° for negative frequencies, with unchanged magnitude
H(ω) = −j·sgn(ω): magnitude 1, phase −90° for ω > 0 and +90° for ω < 0.
25. The Hilbert transform of cos(ω₀t) is:
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Shift the cosine by −90°.
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Answer: A. sin(ω₀t)
A −90° shift of cos(ω₀t) gives cos(ω₀t − 90°) = sin(ω₀t).
26. According to the Wiener–Khinchin theorem, the power spectral density of a wide-sense stationary signal is:
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It links a time-domain correlation to a frequency-domain density.
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Answer: B. The Fourier transform of its autocorrelation function
Sₓ(f) = F{Rₓ(τ)} for a WSS process (Wiener–Khinchin).
27. The energy spectral density of x(t) = e^(−2t)·u(t) at ω = 0 is:
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ESD is the squared magnitude of the Fourier transform.
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Answer: B. 0.25
Ψ(ω) = |X(ω)|² = 1/(4 + ω²); at ω = 0 it is 1/4 = 0.25.
28. The DTFT X(e^jω) of any discrete-time sequence is always:
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Consider e^(−jωn) when ω increases by 2π.
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Answer: C. Periodic in ω with period 2π
Because e^(−j(ω+2π)n) = e^(−jωn) for integer n, X(e^jω) repeats every 2π.
29. For x[n] = (0.5)ⁿ·u[n], the value of its DTFT at ω = 0 is:
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At ω = 0 the DTFT is simply the sum of all samples.
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Answer: C. 2
X(e^jω) = 1/(1 − 0.5e^(−jω)); at ω = 0 it is 1/(1 − 0.5) = 2 (also the sum of the samples).
30. A periodic impulse train x[n] with period N = 4 has x[0] = 1 and x[1] = x[2] = x[3] = 0 in each period. Its DTFS coefficients aₖ (with the 1/N factor in the analysis equation) are:
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Only the n = 0 term survives in the sum.
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Answer: C. aₖ = 1/4 for every k
aₖ = (1/N) Σ x[n]e^(−j2πkn/N) = (1/4)·1 = 1/4 for k = 0,1,2,3.
31. A discrete-time periodic signal with fundamental period N has how many distinct Fourier series coefficients?
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Discrete-time complex exponentials repeat after N harmonics.
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Answer: A. N
The DTFS coefficients are themselves periodic with period N, so only N are distinct.
8.2 Linear time invariant system
30 questions · AExE0802
32. In ideal (impulse-train) sampling at rate fs, the spectrum of the sampled signal consists of:
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Multiplication in time is convolution in frequency.
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Answer: B. The original spectrum repeated at every integer multiple of fs, scaled by 1/Ts
Multiplying by an impulse train convolves the spectrum with an impulse train in frequency, giving copies at k·fs with weight 1/Ts.
33. What is the Nyquist rate for x(t) = cos(400πt) + sin(1000πt)?
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Convert each ω to f first and keep the largest.
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Answer: A. 1000 Hz
The components are at 200 Hz and 500 Hz; the Nyquist rate is twice the highest frequency, 2 × 500 = 1000 Hz.
34. A speech signal band-limited to 3.4 kHz is sampled at its Nyquist rate. The maximum sampling interval is approximately:
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Sampling interval is the reciprocal of the sampling rate.
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Answer: C. 147 µs
Nyquist rate = 6.8 kHz, so Ts = 1/6800 ≈ 147 µs.
35. A 7 kHz sinusoid is sampled at 10 kHz without an anti-aliasing filter. After ideal reconstruction (cutoff fs/2), the output frequency is:
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Fold the frequency about fs/2.
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Answer: C. 3 kHz
7 kHz exceeds fs/2 = 5 kHz, so it aliases to |7 − 10| = 3 kHz.
36. A signal x(t) is band-limited to 3 kHz. The minimum sampling rate for y(t) = x²(t) is:
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What does multiplication in time do to the bandwidth?
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Answer: A. 12 kHz
Squaring convolves the spectrum with itself, doubling the bandwidth to 6 kHz; Nyquist rate = 12 kHz.
37. The main purpose of an anti-aliasing filter placed before an A/D converter is to:
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It acts before sampling, not after.
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Answer: C. Limit the input bandwidth to below half the sampling rate
An analog low-pass filter before sampling removes components above fs/2 that would otherwise alias.
38. A system is time-invariant if:
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Compare the response to a delayed input with the delayed response.
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Answer: B. A time shift of the input causes an identical time shift of the output
Time invariance: x(t − t₀) → y(t − t₀) for every t₀.
39. For a continuous-time LTI system, which condition guarantees BIBO stability?
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The test uses the absolute value of h(t).
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Answer: C. ∫|h(t)| dt over all t is finite
An LTI system is BIBO stable if and only if its impulse response is absolutely integrable.
40. An LTI system has h[n] = (0.5)ⁿ·u[n]. Which statement is correct?
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Sum the geometric series of |h[n]|.
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Answer: A. It is causal and stable, since Σ|h[n]| = 2
h[n] = 0 for n < 0 (causal) and Σ(0.5)ⁿ = 1/(1 − 0.5) = 2 is finite (stable).
41. For stability of a causal LTI system, all poles of the transfer function must lie:
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The jω-axis maps onto the unit circle.
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Answer: D. In the left half of the s-plane for continuous time, and inside the unit circle of the z-plane for discrete time
Causal CT stability needs Re(poles) < 0; causal DT stability needs |poles| < 1.
42. Two LTI systems with impulse responses h₁(t) and h₂(t) are connected in cascade. The overall impulse response is:
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In frequency it is the product of H₁ and H₂.
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Answer: D. h₁(t) * h₂(t)
A cascade of LTI systems has the convolution of the impulse responses (product of frequency responses).
43. An LTI system with impulse response h(t) = K·δ(t) is:
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Does the output depend on past input values?
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Answer: D. Memoryless
The output y(t) = K·x(t) depends only on the present input, so the system has no memory.
44. The complex exponential e^(jω₀t) applied to an LTI system produces the output H(jω₀)e^(jω₀t). For this reason complex exponentials are called:
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The input shape is preserved; only a complex scale factor appears.
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Answer: C. Eigenfunctions of LTI systems
The output is the same exponential scaled by the eigenvalue H(jω₀), so e^(jω₀t) is an eigenfunction.
45. For the system y[n] = 0.5(x[n] + x[n − 1]), what is |H(e^jω)| at ω = π/2?
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Factor out e^(−jω/2) to get a cosine.
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Answer: C. 0.707
H(e^jω) = 0.5(1 + e^(−jω)) ⇒ |H| = |cos(ω/2)| = cos(π/4) ≈ 0.707.
46. The system y[n] = 0.5(x[n] + x[n − 1]) behaves as a:
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Evaluate the gain at DC and at the highest digital frequency.
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Answer: B. Low-pass filter with a null at ω = π
|H| = |cos(ω/2)| equals 1 at ω = 0 and 0 at ω = π, so it passes low frequencies.
47. An RC low-pass filter has R = 1 kΩ and C = 1 µF. Its 3 dB cutoff frequency is about:
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RC is the time constant; convert rad/s to Hz.
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Answer: C. 159 Hz
fc = 1/(2πRC) = 1/(2π × 10⁻³) ≈ 159 Hz.
48. A signal cos(t) is applied to an LTI system with H(jω) = 1/(1 + jω). The output is:
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Evaluate H at the input frequency.
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Answer: D. 0.707 cos(t − 45°)
At ω = 1: |H| = 1/√2 ≈ 0.707 and ∠H = −45°, so y(t) = 0.707 cos(t − 45°).
49. For distortionless transmission through an LTI system, the frequency response must have:
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The output may be scaled and delayed, nothing else.
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Answer: D. Constant magnitude and phase varying linearly with frequency
y(t) = K·x(t − t_d) requires H(ω) = K e^(−jωt_d): flat magnitude, linear phase.
50. A system has a phase response θ(ω) = −0.002ω rad over its passband. The delay experienced by signals is:
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Differentiate the phase with respect to ω.
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Answer: B. 2 ms
Group delay = −dθ/dω = 0.002 s = 2 ms.
51. The impulse response of an ideal low-pass filter with cutoff ωc (zero phase) is:
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Use duality between rectangle and sinc.
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Answer: B. h(t) = sin(ωc t)/(πt), a sinc pulse extending over all time
The inverse Fourier transform of a rectangular frequency response is a sinc function, extending from −∞ to ∞.
52. An ideal low-pass filter has cutoff fc = 1 kHz and unity passband gain. The peak value of its impulse response h(0) is:
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h(0) equals the area under the frequency response.
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Answer: D. 2000
h(t) = 2fc·sinc(2fc t), so h(0) = 2fc = 2000 (equal to the area under H(f)).
53. An ideal low-pass filter is not physically realizable because:
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Look at when the impulse response starts.
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Answer: A. Its impulse response is non-zero for t < 0, so it is non-causal
The sinc impulse response exists for negative time; a causal filter cannot have a brick-wall magnitude (Paley–Wiener criterion).
54. The step response of an ideal low-pass filter shows:
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Integrating a sinc gives the sine-integral function.
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Answer: B. An overshoot of about 9% with ripples, and a rise time inversely proportional to the bandwidth
The step response is the sine-integral function; Gibbs effect gives ≈9% overshoot, and wider bandwidth gives faster rise.
55. An LTI system has h(t) = e^(−t)·u(t). Its unit step response at t = 1 s is approximately:
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The step response is the running integral of the impulse response.
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Answer: A. 0.632
s(t) = ∫₀^t e^(−τ)dτ = 1 − e^(−t); s(1) = 1 − 0.368 = 0.632.
56. The impulse response of the system y[n] = x[n] − x[n − 2] is:
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Replace x[n] by δ[n].
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Answer: B. {1, 0, −1}
Setting x[n] = δ[n]: h[n] = δ[n] − δ[n − 2] = {1, 0, −1} starting at n = 0.
57. Which of the following discrete-time systems is time-variant?
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Look for a coefficient that depends explicitly on time.
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Answer: B. y[n] = n·x[n]
In y[n] = n·x[n] the coefficient depends on n, so a shifted input does not give a shifted output.
58. Which basic elements are required to implement a discrete-time LTI system from its difference equation?
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Look at the operations appearing in a difference equation.
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Answer: A. Adders, constant multipliers and unit delays
Any linear constant-coefficient difference equation can be realised with adders, gain multipliers and z⁻¹ delay elements.
59. A causal system is described by y[n] = a₁y[n−1] + a₂y[n−2] + b₀x[n] + b₁x[n−1] + b₂x[n−2]. What is the minimum number of delay elements needed (canonic form)?
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The canonic structure uses as many delays as the system order.
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Answer: D. 2
Direct form II (canonic) shares delays between the feedback and feedforward parts, needing max(N, M) = 2 delays.
60. To reconstruct a band-limited signal (bandwidth W) from its samples taken at fs > 2W, the ideal reconstruction filter is a low-pass filter whose cutoff lies:
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Sketch the baseband spectrum and its first image.
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Answer: A. Between W and fs − W
The cutoff must pass the baseband spectrum (up to W) and reject the first image starting at fs − W.
61. Flat-top sampling, compared with ideal sampling, introduces:
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The holding pulse acts like a filter.
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Answer: A. Aperture effect: high-frequency attenuation caused by the sinc-shaped spectrum of the holding pulse
Holding each sample for a pulse width τ multiplies the spectrum by a sinc of τ, attenuating high frequencies (aperture effect); an equaliser can correct it.
8.3 Z-Transform and discrete Fourier transform
30 questions · AExE0803
62. The (bilateral) z-transform of a sequence x[n] is defined as:
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The power of z carries a negative sign.
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Answer: A. X(z) = Σ x[n]·z⁻ⁿ summed over all n
The bilateral z-transform is Σ from −∞ to ∞ of x[n]z⁻ⁿ; the DTFT is its value on the unit circle.
63. The z-transform of x[n] = aⁿ·u[n] and its region of convergence are:
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It is a geometric series in a z⁻¹.
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Answer: D. 1/(1 − a z⁻¹), |z| > |a|
Σ (a z⁻¹)ⁿ converges to 1/(1 − a z⁻¹) when |a z⁻¹| < 1, i.e. |z| > |a|.
64. Both aⁿu[n] and −aⁿu[−n−1] have the same algebraic z-transform 1/(1 − a z⁻¹). They are distinguished by:
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The expression is identical, so something else must differ.
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Answer: C. Their regions of convergence
The first has ROC |z| > |a| (right-sided), the second |z| < |a| (left-sided); X(z) alone is not unique without the ROC.
65. For a finite-duration causal sequence such as x[n] = {1, 2, 3} (n = 0, 1, 2), the ROC is:
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Where do terms in z⁻¹ become infinite?
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Answer: D. The entire z-plane except z = 0
X(z) = 1 + 2z⁻¹ + 3z⁻² is finite everywhere except at z = 0 where the negative powers blow up.
66. Which statement about the region of convergence of a z-transform is correct?
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What is the value of X(z) at a pole?
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Answer: B. The ROC cannot contain any pole
At a pole X(z) is infinite, so poles bound the ROC; a two-sided sequence has an annular ROC.
67. A system has H(z) with poles at z = 0.5 and z = 2. If the system is causal, then it is:
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Causality fixes the ROC; stability needs the unit circle inside it.
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Answer: C. Unstable, because its ROC |z| > 2 does not include the unit circle
Causal ⇒ ROC outside the outermost pole (|z| > 2), which excludes |z| = 1, so it is unstable.
68. The z-transform of x[n − k] (k > 0, causal x) is:
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z⁻¹ is the unit delay operator.
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Answer: D. z⁻ᵏ X(z)
Time-shift property: a delay of k samples multiplies X(z) by z⁻ᵏ.
69. The convolution y[n] = x[n] * h[n] corresponds in the z-domain to:
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Same idea as the Laplace and Fourier convolution properties.
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Answer: D. Y(z) = X(z)·H(z)
The convolution property of the z-transform: convolution in time becomes multiplication in z.
70. The inverse z-transform of X(z) = 1/(1 − 0.5z⁻¹), |z| > 0.5, evaluated at n = 2 is:
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Identify the standard pair aⁿu[n].
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Answer: A. 0.25
x[n] = (0.5)ⁿu[n], so x[2] = 0.25.
71. X(z) = z / [(z − 1)(z − 0.5)] with ROC |z| > 1. Using the final value theorem, lim x[n] as n → ∞ is:
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Multiply X(z) by (z − 1) and let z → 1.
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Answer: A. 2
x(∞) = lim(z→1) (z − 1)X(z) = 1/(1 − 0.5) = 2. Partial fractions give x[n] = 2 − 2(0.5)ⁿ for n ≥ 0, which tends to 2.
72. X(z) = (2z² + z)/(z² − 0.5z + 0.1) is the z-transform of a causal sequence. The initial value x[0] is:
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Let z grow very large and compare the leading terms.
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Answer: B. 2
Initial value theorem: x[0] = lim(z→∞) X(z) = 2/1 = 2.
73. By Parseval's theorem, the energy of x[n] = (0.5)ⁿ·u[n] is:
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Square each sample and sum the geometric series.
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Answer: B. 4/3
Σ|x[n]|² = Σ(0.25)ⁿ = 1/(1 − 0.25) = 4/3; the same value results from (1/2π)∫|X(e^jω)|² dω.
74. A stable causal system H(z) = (1 + z⁻¹)/2 is driven by cos(πn/3). The steady-state output amplitude is:
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Evaluate H(z) on the unit circle at z = e^(jπ/3).
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Answer: C. 0.866
|H(e^jω)| = |cos(ω/2)|; at ω = π/3, cos(π/6) ≈ 0.866.
75. When a sinusoid is suddenly applied at n = 0 to a stable causal system H(z), the total response consists of:
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Separate the response into terms from the system poles and from the input poles.
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Answer: A. A transient part from the poles of H(z) that decays to zero, plus a steady-state sinusoid scaled by H(e^jω₀)
For a stable system the system-pole terms die out, leaving |H(e^jω₀)|cos(ω₀n + ∠H(e^jω₀)).
76. What is the DC gain of H(z) = 1/(1 − 0.8z⁻¹)?
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DC (ω = 0) maps to z = 1.
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Answer: B. 5
DC corresponds to z = 1: H(1) = 1/(1 − 0.8) = 5.
77. A filter has a single zero at z = −1 and a single pole at z = 0.9. It acts as a:
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z = 1 is ω = 0 and z = −1 is ω = π.
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Answer: B. Low-pass filter
The pole near z = 1 boosts low frequencies and the zero at z = −1 (ω = π) kills the highest frequency, giving a low-pass response.
78. If the DTFT of a finite sequence of length L is sampled at N equally spaced frequencies with N < L, the inverse DFT of those samples gives:
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Sampling in one domain causes periodic repetition in the other.
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Answer: B. A time-aliased version of x[n]
Frequency-domain sampling makes the time sequence periodic with period N; if N < L the periodic copies overlap (time aliasing).
79. The N-point DFT X[k] of a finite sequence is related to its z-transform by:
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The DFT samples lie on the unit circle.
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Answer: B. X[k] = X(z) evaluated at z = e^(j2πk/N)
The DFT samples the z-transform at N equally spaced points on the unit circle.
80. The 4-point DFT of x[n] = {1, 1, 1, 1} is:
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A constant has energy only at DC.
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Answer: D. {4, 0, 0, 0}
X[0] = sum of samples = 4; for k ≠ 0 the complex exponentials sum to zero.
81. For x[n] = {1, 2, 3, 4}, the 4-point DFT value X[1] is:
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For N = 4, W₄ = −j.
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Answer: C. −2 + 2j
X[1] = Σ x[n](−j)ⁿ = 1 − 2j − 3 + 4j = −2 + 2j.
82. A signal sampled at 8 kHz is analysed with a 256-point DFT. The frequency spacing between DFT bins is:
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Divide the sampling rate by the number of points.
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Answer: A. 31.25 Hz
Δf = fs/N = 8000/256 = 31.25 Hz.
83. The 2-point circular convolution of x[n] = {1, 2} and h[n] = {3, 4} is:
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Indices wrap modulo 2.
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Answer: C. {11, 10}
y[0] = 1·3 + 2·4 = 11 and y[1] = 1·4 + 2·3 = 10.
84. Two sequences of lengths 50 and 30 are to be linearly convolved using DFTs. The minimum DFT length that avoids wrap-around error is:
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Find the length of the linear convolution result.
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Answer: B. 79
Circular convolution equals linear convolution when N ≥ L + M − 1 = 50 + 30 − 1 = 79.
85. Multiplying the N-point DFTs of two sequences and taking the inverse DFT gives their:
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DFT treats sequences as one period of a periodic signal.
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Answer: A. N-point circular convolution
The DFT product property: X₁[k]X₂[k] ↔ x₁[n] ⊛ x₂[n] (circular convolution modulo N).
86. x[n] is real and its 8-point DFT has X[1] = 2 + 3j. Then X[7] equals:
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Real sequences have conjugate-symmetric DFTs.
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Answer: A. 2 − 3j
For real x[n], X[N − k] = X*[k], so X[7] = X*[1] = 2 − 3j.
87. A circular shift x[(n − m) mod N] in time corresponds in the DFT domain to:
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A shift in one domain is a phase factor in the other.
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Answer: C. Multiplying X[k] by e^(−j2πkm/N)
Circular time shift ↔ multiplication by the linear phase factor W_N^(km) = e^(−j2πkm/N).
88. For x[n] = {1, 2, 3, 4}, Σ|X[k]|² over its 4-point DFT equals:
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Mind the 1/N factor in the DFT form of Parseval's relation.
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Answer: C. 120
Parseval for the DFT: Σ|x[n]|² = (1/N)Σ|X[k]|², so Σ|X[k]|² = 4 × 30 = 120.
89. The N-point DFT of the unit impulse δ[n] is:
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Only the n = 0 term contributes.
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Answer: A. 1 for every k
X[k] = Σ δ[n]W^(kn) = W⁰ = 1 for all k.
90. The N-point DFT X[k], viewed as a function of the integer k, is:
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Replace k by k + N in the twiddle factor.
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Answer: D. Periodic with period N
W_N^((k+N)n) = W_N^(kn), so X[k + N] = X[k].
91. Compared with the DTFT, the DFT of a finite sequence of length L (N ≥ L) is:
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The DFT is discrete in frequency.
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Answer: D. N equally spaced samples of the DTFT over one period
X[k] = X(e^jω) at ω = 2πk/N, k = 0 … N − 1.
8.4 Implementation of discrete-time system
30 questions · AExE0804
92. An IIR filter has a numerator of order 4 and a denominator of order 4. How many delay elements are needed in Direct Form I and in Direct Form II respectively?
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Which form shares one delay line between the zeros and poles?
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Answer: C. 8 and 4
Direct Form I uses separate delay lines for input and output (M + N = 8); Direct Form II shares them (max(M, N) = 4).
93. Direct Form II realization of an IIR filter is called canonic because:
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Count the delays relative to the filter order.
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Answer: C. It uses the minimum possible number of delay elements
A structure is canonic when the number of delays equals the order of the difference equation, which Direct Form II achieves.
94. The transposed form of a structure is obtained by:
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It is a flow-graph manipulation that keeps H(z) the same.
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Answer: D. Reversing all branch directions, swapping input and output, and interchanging adders and branch nodes
By the transposition (flow-graph reversal) theorem, these operations leave the transfer function unchanged.
95. A direct-form FIR filter of length M = 21 has symmetric coefficients h[n] = h[20 − n]. Exploiting the symmetry, the number of multipliers needed is:
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Group equal coefficients together and count the unique values.
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Answer: D. 11
Pairs of equal coefficients share one multiplier: (M + 1)/2 = 11 for odd M (10 pairs plus the centre tap).
96. A 6th-order IIR filter is realised in cascade form using second-order sections. How many sections are needed?
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Each section handles a pair of poles.
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Answer: A. 3
Each biquad realises two poles (and two zeros), so 6/2 = 3 sections.
97. The parallel-form realization of an IIR filter is obtained from:
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Parallel means the outputs are added.
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Answer: C. A partial-fraction expansion of H(z)
Partial fractions express H(z) as a sum of first/second-order terms, which are realised in parallel; factoring gives the cascade form.
98. A two-stage FIR lattice has reflection coefficients K₁ = 0.5 and K₂ = 0.25. The direct-form polynomial is A₂(z) = 1 + a₁z⁻¹ + a₂z⁻². What are a₁ and a₂?
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The last coefficient always equals the last reflection coefficient.
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Answer: B. a₁ = 0.625, a₂ = 0.25
a₂ = K₂ = 0.25 and a₁ = K₁(1 + K₂) = 0.5 × 1.25 = 0.625.
99. An FIR filter is A₂(z) = 1 + 0.6z⁻¹ + 0.2z⁻². The lattice reflection coefficients are:
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Use the step-down (backward) recursion.
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Answer: B. K₂ = 0.2, K₁ = 0.5
K₂ = a₂ = 0.2; stepping down, K₁ = a₁/(1 + K₂) = 0.6/1.2 = 0.5.
100. For an all-pole IIR lattice filter, the system is stable if and only if:
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This is a quick stability check without finding roots.
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Answer: A. All reflection coefficients satisfy |Kₘ| < 1
|Kₘ| < 1 for all m is equivalent to all roots of A_N(z) lying inside the unit circle.
101. A filter H(z) = B(z)/A(z) having both poles and zeros is realised in lattice form by:
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One part handles the poles, another part the zeros.
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Answer: D. A lattice–ladder structure: the all-pole lattice for A(z) with ladder (tap) coefficients forming B(z)
The lattice part realises the poles; the ladder coefficients combine the backward signals to realise the numerator.
102. Which statement correctly compares FIR and IIR digital filters?
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Think about feedback and phase.
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Answer: D. FIR filters can have exactly linear phase and are always stable; IIR filters need fewer coefficients for a sharp response
FIR has no feedback (no poles except at the origin) and symmetric taps give linear phase; IIR uses feedback to achieve sharp responses with low order.
103. In 8-bit two's-complement integer representation, the range of numbers is:
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There is one more negative number than positive.
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Answer: C. −128 to +127
b-bit two's complement covers −2^(b−1) to 2^(b−1) − 1 = −128 to 127.
104. In the Q15 fixed-point format (1 sign bit, 15 fractional bits), the smallest step (resolution) is about:
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Count the fractional bits.
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Answer: B. 3.05 × 10⁻⁵
Resolution = 2⁻¹⁵ ≈ 3.05 × 10⁻⁵.
105. The decimal fraction 0.625 in binary fixed-point form is:
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Write it as a sum of powers of 1/2.
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Answer: A. 0.101
0.625 = 0.5 + 0.125 = 2⁻¹ + 2⁻³ = 0.101₂.
106. Using 4 bits (1 sign bit + 3 fractional bits) in two's complement, −0.375 is represented as:
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Invert the bits of +0.375 and add one LSB.
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Answer: A. 1.101
+0.375 = 0.011; inverting gives 1.100 and adding 1 LSB gives 1.101.
107. The main advantage of floating-point representation over fixed-point in a DSP system is:
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What does the exponent field provide?
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Answer: A. A much larger dynamic range for the same word length
The exponent lets floating-point represent very large and very small values; fixed-point hardware is simpler but has limited range.
108. In IEEE 754 single-precision floating point, the 32 bits are divided as:
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The double-precision format uses an 11-bit exponent.
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Answer: D. 1 sign, 8 exponent, 23 mantissa (fraction)
Single precision: 1 sign bit, 8-bit biased exponent, 23-bit fraction (double uses 1/11/52).
109. When a value is quantized by rounding with step Δ, the quantization error e lies in the range:
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Rounding picks the nearest level.
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Answer: B. −Δ/2 < e ≤ Δ/2
Rounding chooses the nearest level, so the error never exceeds half a step in magnitude.
110. For truncation of a two's-complement number with step Δ, the error e lies in:
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In two's complement, dropping bits never increases the value.
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Answer: B. −Δ < e ≤ 0
Two's-complement truncation always moves the value downward (toward −∞), so the error is between −Δ and 0 regardless of sign.
111. Rounding noise is modelled as uniformly distributed with step Δ = 2⁻⁷. Its variance is approximately:
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Use the variance of a uniform distribution of width Δ.
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Answer: C. 5.09 × 10⁻⁶
σ² = Δ²/12 = 2⁻¹⁴/12 ≈ 5.09 × 10⁻⁶.
112. For a full-scale sinusoid, the signal-to-quantization-noise ratio of a 12-bit quantizer is about:
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Each bit is worth about 6 dB.
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Answer: C. 74 dB
SQNR ≈ 6.02B + 1.76 = 6.02 × 12 + 1.76 ≈ 74 dB.
113. Multiplying two b-bit fixed-point fractions produces a result with:
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Think of multiplying two 3-digit decimals.
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Answer: C. About 2b bits, which must be rounded or truncated back to b bits
The product of two b-bit numbers needs about 2b bits; reducing it to the register length introduces product round-off noise.
114. The pole of H(z) = 1/(1 − 0.95z⁻¹) is implemented with the coefficient rounded to 3 fractional bits. The pole moves to:
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Multiply by 2³, round to the nearest integer, divide back.
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Answer: B. 1.0, making the filter marginally stable
0.95 × 8 = 7.6 rounds to 8, so the coefficient becomes 8/8 = 1.0 and the pole lies on the unit circle.
115. Why are high-order IIR filters usually implemented as cascades of second-order sections rather than in direct form?
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Think about how polynomial roots respond to small coefficient changes.
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Answer: A. Pole positions of a high-order direct form are very sensitive to coefficient quantization
Each root of a high-order polynomial depends on all coefficients, so small quantization errors can move clustered poles a lot, even outside the unit circle; second-order sections localise the effect.
116. Pole sensitivity to coefficient quantization in a direct-form IIR filter is greatest when:
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Look at the product of pole-distance terms in the sensitivity formula.
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Answer: A. The poles are tightly clustered, as in a narrow-band filter
Sensitivity is inversely proportional to the distances between poles, so closely spaced poles move most.
117. Zero-input limit cycles (sustained oscillations with no input) can occur in:
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Limit cycles need a feedback path.
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Answer: D. Recursive (IIR) filters implemented with finite-precision arithmetic
Rounding inside a feedback loop can keep the state from decaying; FIR filters have no feedback and cannot sustain limit cycles.
118. To prevent overflow in a fixed-point filter, a common technique is to:
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Keep the internal signals within the representable range.
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Answer: B. Scale the input (or internal signals) and/or use saturation arithmetic
Scaling keeps internal node values within range; saturation clips instead of wrapping around, avoiding large errors and overflow oscillations.
119. Quantizing the coefficients of an FIR filter can alter its frequency response, but it can never:
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Where are the poles of an FIR filter?
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Answer: A. Make the filter unstable
An FIR filter has no poles other than at z = 0, so quantized coefficients move only the zeros; the filter remains BIBO stable.
120. The number of multiplications per output sample for a second-order IIR section (biquad) in direct form II is:
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Count numerator and denominator coefficients (a₀ = 1).
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Answer: B. 5
A biquad has three numerator coefficients (b₀, b₁, b₂) and two denominator coefficients (a₁, a₂): 5 multiplications.
121. Which of the following is an advantage of lattice structures in digital filter implementation?
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Think about what the reflection coefficients reveal.
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Answer: D. Low sensitivity to coefficient quantization and an easy stability check through |Kₘ| < 1
Lattice structures are modular and robust to quantization; stability is checked directly from the reflection coefficients. They usually need more multipliers.
8.5 IIR filter design and FIR filter design
30 questions · AExE0805
122. In the classical approach to IIR filter design, a digital filter is obtained by:
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IIR design borrows from mature analog filter theory.
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Answer: D. Designing an analog prototype (e.g. Butterworth or Chebyshev) and mapping it to the z-domain
IIR design commonly converts a well-known analog filter to digital via impulse invariance or the bilinear transformation; the other options are FIR methods.
123. The magnitude response of a Butterworth low-pass filter is:
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No ripple anywhere.
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Answer: A. Maximally flat in the passband and monotonic everywhere
|H|² = 1/(1 + (Ω/Ωc)^(2N)) has all derivatives zero at Ω = 0 and decreases monotonically.
124. At its cutoff frequency Ωc, the gain of a Butterworth filter of any order is:
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Substitute Ω = Ωc in |H|².
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Answer: C. −3 dB
At Ω = Ωc, |H|² = 1/2, i.e. 10 log₁₀(0.5) ≈ −3 dB regardless of N.
125. A 4th-order Butterworth low-pass filter has an asymptotic stopband roll-off of:
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Each pole contributes 20 dB/decade.
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Answer: B. 80 dB/decade
Roll-off = 20N dB/decade = 20 × 4 = 80 dB/decade (≈ 24 dB/octave).
126. What is the minimum order of a Butterworth low-pass filter that gives at least 30 dB attenuation at twice the cutoff (3 dB) frequency?
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Set 10 log₁₀(1 + 2^(2N)) ≥ 30 and solve for N.
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Answer: A. 5
Need 1 + 2^(2N) ≥ 10³ ⇒ N ≥ log₁₀(999)/(2 log₁₀2) ≈ 4.98, so N = 5.
127. Which analog filter type gives the lowest order for a given set of passband ripple, stopband attenuation and transition width?
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The one with ripple in both bands.
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Answer: A. Elliptic (Cauer)
Elliptic filters have equiripple behaviour in both bands, giving the sharpest transition for a given order.
128. A Chebyshev Type I filter has ripple parameter ε = 0.5. The passband ripple is about:
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Passband gain oscillates between 1 and 1/√(1 + ε²).
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Answer: B. 0.97 dB
Ripple = 10 log₁₀(1 + ε²) = 10 log₁₀(1.25) ≈ 0.97 dB.
129. Which statement about Chebyshev filters is correct?
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Inverse Chebyshev moves the ripple to the other band.
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Answer: B. Type I has equiripple passband and monotonic stopband; Type II has monotonic passband and equiripple stopband
Type I (Chebyshev) ripples in the passband; Type II (inverse Chebyshev) ripples in the stopband.
130. In the impulse-invariance method, an analog pole at s = pₖ maps to a digital pole at:
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Sample the exponential e^(pₖt) at t = nT.
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Answer: B. z = e^(pₖT)
Sampling hₐ(t) = e^(pₖt) at t = nT gives (e^(pₖT))ⁿ, i.e. a pole at z = e^(pₖT).
131. Using impulse invariance with T = 0.1 s, the analog filter H(s) = 1/(s + 2) gives a digital pole at approximately:
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Use z = e^(pT) with p = −2.
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Answer: A. 0.819
z = e^(−2 × 0.1) = e^(−0.2) ≈ 0.819.
132. The impulse-invariance method is unsuitable for designing high-pass and band-stop filters because:
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Sampling a non-band-limited response causes what?
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Answer: A. The frequency response of the sampled impulse response suffers from aliasing
Impulse invariance samples hₐ(t); since high-pass responses are not band-limited, the spectral copies overlap (aliasing).
133. The bilinear transformation used in IIR design is:
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It comes from the trapezoidal rule of integration.
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Answer: A. s = (2/T)·(1 − z⁻¹)/(1 + z⁻¹)
The bilinear (Tustin) map is s = (2/T)(1 − z⁻¹)/(1 + z⁻¹); it maps the jΩ axis once onto the unit circle.
134. The main property of the bilinear transformation that impulse invariance lacks is:
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Think about how the whole analog frequency axis fits into −π to π.
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Answer: B. It avoids aliasing by mapping the entire jΩ axis exactly once onto the unit circle
The one-to-one mapping prevents aliasing, at the cost of non-linear frequency warping ω = 2 tan⁻¹(ΩT/2).
135. A digital low-pass filter with cutoff 2 kHz at fs = 8 kHz is to be designed by the bilinear transformation (T = 1/fs). The pre-warped analog cutoff is:
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Use Ω = (2/T) tan(ω/2).
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Answer: D. 16000 rad/s
ω = 2π(2000/8000) = π/2; Ω = (2/T)tan(ω/2) = 16000 × tan(π/4) = 16000 rad/s.
136. For a sampling frequency of 8 kHz, an analog frequency of 1 kHz corresponds to a normalised digital frequency of:
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fs corresponds to 2π rad/sample.
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Answer: B. π/4 rad/sample
ω = 2πf/fs = 2π × 1000/8000 = π/4 rad/sample.
137. Which statement about IIR and FIR filter design is correct?
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Weigh order against phase linearity.
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Answer: B. An IIR design generally meets a magnitude specification with a much lower order than an FIR design, but its phase is non-linear
Feedback poles let IIR filters achieve sharp responses with low order; FIR filters can be exactly linear phase and are always stable.
138. In FIR design by Fourier series (window) method, the ideal low-pass impulse response with cutoff ωc and delay α is hd[n] = sin(ωc(n − α))/(π(n − α)). For ωc = π/4, hd[α] equals:
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Use sin(x)/x → 1 as x → 0.
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Answer: D. 0.25
At n = α the limit is ωc/π = 0.25.
139. Truncating the ideal impulse response with a rectangular window causes:
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Abrupt truncation in time causes ripple in frequency.
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Answer: A. Gibbs oscillations, giving a minimum stopband attenuation of only about 21 dB
The rectangular window's large sidelobes produce ripple near the band edge; peak stopband attenuation is about 21 dB regardless of length.
140. Arrange the windows in increasing order of minimum stopband attenuation:
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The rectangular window is the worst.
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Answer: A. Rectangular, Hanning, Hamming, Blackman
Typical values: rectangular ≈ 21 dB, Hanning ≈ 44 dB, Hamming ≈ 53 dB, Blackman ≈ 74 dB.
141. The Hamming window w[n] = 0.54 − 0.46 cos(2πn/(M − 1)) has the value at its end points (n = 0 and n = M − 1) of:
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Put n = 0 into the formula.
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Answer: C. 0.08
At n = 0, cos(0) = 1 so w = 0.54 − 0.46 = 0.08 (unlike the Hanning window, which goes to zero).
142. Using the approximation that the transition width of a Hamming-window design equals its main-lobe width 8π/M, the length M needed for a transition width of 0.1π rad/sample is:
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Solve 8π/M = 0.1π for M.
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Answer: D. 80
M = 8π/Δω = 8π/(0.1π) = 80.
143. The Kaiser window is preferred in many FIR designs because:
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It has an adjustable parameter.
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Answer: C. Its shape parameter β allows a trade-off between main-lobe width and sidelobe level
Varying β adjusts sidelobe attenuation; together with M it meets a given ripple and transition width.
144. A linear-phase FIR filter of length M = 31 has a group delay of:
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The delay is to the centre of symmetry.
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Answer: D. 15 samples
Group delay = (M − 1)/2 = 30/2 = 15 samples.
145. A linear-phase FIR filter with symmetric impulse response of even length (Type II) cannot be used as a:
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Evaluate H(e^jω) at ω = π for an even-length symmetric filter.
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Answer: B. High-pass filter
Type II filters always have a zero at z = −1 (ω = π), forcing the gain to zero at the highest frequency.
146. In the frequency-sampling method of FIR filter design:
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The name says what is sampled.
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Answer: C. Desired frequency-response samples at ωₖ = 2πk/M are specified and h[n] is found by the inverse DFT
The filter is defined by its DFT samples; free transition-band samples can be optimised to improve stopband attenuation.
147. A length M = 5 linear-phase FIR filter is designed by frequency sampling with amplitude samples |H(k)| = 1, 1, 0, 0, 1 for k = 0 … 4. The centre tap h[2] is:
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At n = (M − 1)/2 the phase factors all become 1.
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Answer: D. 0.6
At the centre of symmetry the linear-phase terms cancel, so h[2] = (1/M)Σ|H(k)| = 3/5 = 0.6.
148. The Remez exchange algorithm (Parks–McClellan method) designs FIR filters that:
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Minimax means minimising the worst case.
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Answer: C. Minimise the maximum weighted approximation error, giving equiripple behaviour
It solves a Chebyshev (minimax) approximation problem, producing an optimal equiripple linear-phase filter.
149. The optimality of an equiripple linear-phase FIR filter is characterised by the alternation theorem, which requires the weighted error to have:
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The name of the theorem describes the shape of the error.
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Answer: D. At least L + 2 extremal frequencies with alternating sign and equal magnitude, where L is the degree of the cosine polynomial
The best minimax approximation by an L-th degree polynomial has an error curve alternating at least L + 2 times.
150. Compared with a window-designed FIR filter of the same length, an optimum equiripple (Parks–McClellan) design:
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Think about where the error is concentrated in window designs.
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Answer: C. Achieves a narrower transition band or smaller ripple, because the error is spread evenly over each band
Window designs concentrate the error near band edges; equiripple designs distribute it, making better use of the filter length.
151. A Bessel analog prototype is chosen when the main requirement is:
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It is about delay, not sharpness.
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Answer: C. A maximally flat group delay (nearly linear phase) in the passband
Bessel (Thomson) filters approximate a constant time delay, preserving pulse shape, at the cost of a slow roll-off.
8.6 Digital filter
30 questions · AExE0806
152. For N = 1024, the number of complex multiplications needed by a direct DFT and by a radix-2 FFT are respectively:
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FFT cost is (N/2)·log₂N multiplications.
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Answer: A. 1,048,576 and 5,120
Direct DFT needs N² = 1,048,576; radix-2 FFT needs (N/2)log₂N = 512 × 10 = 5,120.
153. The computational advantage of a 1024-point radix-2 FFT over a direct DFT, measured by complex multiplications, is about:
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Divide N² by (N/2)log₂N.
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Answer: C. 205 times
N²/((N/2)log₂N) = 2N/log₂N = 2048/10 ≈ 205.
154. A 256-point radix-2 FFT has how many stages, and how many butterflies per stage?
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Each butterfly handles two points.
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Answer: C. 8 stages, 128 butterflies per stage
Stages = log₂256 = 8; each stage has N/2 = 128 butterflies (1024 butterflies in total).
155. In a radix-2 decimation-in-time (DIT) FFT with in-place computation, the input and output orders are:
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DIT splits the time sequence.
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Answer: C. Input in bit-reversed order, output in natural order
DIT repeatedly splits the input into even and odd samples, which scrambles the input into bit-reversed order; the output emerges in natural order.
156. In a radix-2 decimation-in-frequency (DIF) FFT with in-place computation:
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DIF is the mirror image of DIT.
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Answer: A. The input is in natural order and the output is in bit-reversed order
DIF splits the output (frequency) samples into even and odd indices, so the output appears bit-reversed.
157. For an 8-point FFT, input sample x[3] is placed at which position after bit reversal?
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Write the index with log₂N bits and reverse them.
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Answer: B. x[6]
3 = 011₂; reversing the 3 bits gives 110₂ = 6.
158. For a 16-point FFT, the bit-reversed position of index 1 is:
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Use 4 bits because 16 = 2⁴.
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Answer: D. 8
1 = 0001₂ (4 bits); reversed it is 1000₂ = 8.
159. A radix-2 butterfly computes A = a + W·b and B = a − W·b. It requires:
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The product W·b is shared.
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Answer: B. One complex multiplication and two complex additions
W·b is computed once and then added to and subtracted from a.
160. The twiddle factor W₈² (where W_N = e^(−j2π/N)) equals:
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Compute the angle 2π·2/8.
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Answer: B. −j
W₈² = e^(−j2π·2/8) = e^(−jπ/2) = −j.
161. The symmetry property of the twiddle factor used to halve the multiplications in the FFT is:
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What is W_N^(N/2)?
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Answer: D. W_N^(k + N/2) = −W_N^k
W_N^(N/2) = e^(−jπ) = −1, so W_N^(k+N/2) = −W_N^k; periodicity gives W_N^(k+N) = W_N^k.
162. The term in-place computation in an FFT means:
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It is about memory.
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Answer: D. The outputs of each butterfly are stored in the same memory locations as its inputs
Since each butterfly's inputs are not needed again, the results overwrite them, so only N storage locations are required.
163. Appending zeros to a sequence before computing its DFT (zero padding):
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Does zero padding add any new information about the signal?
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Answer: A. Gives more closely spaced samples of the same DTFT but does not improve the true frequency resolution
Zero padding interpolates the spectrum; resolution depends on the actual data length (record duration).
164. A signal sampled at 10 kHz must be analysed with a frequency spacing of at most 5 Hz using a radix-2 FFT. The minimum FFT length is:
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Compute fs/Δf, then round up to a power of 2.
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Answer: D. 2048
N ≥ fs/Δf = 10000/5 = 2000; the next power of 2 is 2048.
165. In the overlap-save method with a 64-point FFT and an FIR filter of length 17, the number of new valid output samples obtained from each block is:
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M − 1 outputs per block are corrupted by circular wrap-around.
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Answer: C. 48
Each block yields N − M + 1 = 64 − 17 + 1 = 48 valid outputs; the first M − 1 = 16 are discarded.
166. In the overlap-add method, input blocks of L = 100 samples are filtered by an FIR filter of length M = 29 using a radix-2 FFT. The minimum FFT size is:
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Linear convolution of each block must fit in the FFT.
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Answer: B. 128
Each block convolution has length L + M − 1 = 128, which happens to be a power of 2.
167. In the overlap-save method of fast convolution, from each output block one must:
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Which samples are corrupted by circular convolution?
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Answer: C. Discard the first M − 1 samples, where M is the filter length
Overlap-save overlaps the inputs and discards the aliased outputs; overlap-add instead adds the overlapping output tails.
168. The Goertzel algorithm is preferred over the FFT when:
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Telephone keypad tone detection uses it.
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Answer: B. Only a few DFT values are needed, as in DTMF tone detection
Goertzel computes individual DFT bins with a second-order recursive filter; for a few bins it is cheaper than a full FFT.
169. The inverse DFT can be computed with a forward FFT routine by:
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The IDFT differs from the DFT only in the sign of the exponent and a 1/N factor.
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Answer: D. Conjugating X[k], applying the FFT, conjugating the result and dividing by N
x[n] = (1/N)[FFT{X*[k]}]*, so no separate IFFT routine is needed.
170. Two real N-point sequences can have their DFTs computed efficiently by:
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A complex sequence has two real parts to carry.
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Answer: A. Forming one complex sequence x₁[n] + j x₂[n] and computing a single N-point FFT
The DFTs are recovered using conjugate-symmetry: X₁[k] = (Y[k] + Y*[N−k])/2 and X₂[k] = (Y[k] − Y*[N−k])/(2j).
171. Spectral leakage in FFT-based spectrum analysis is caused by:
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Truncating a signal is the same as multiplying by a window.
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Answer: B. Analysing a finite record that does not contain an integer number of periods, and is reduced by windowing
Truncation is multiplication by a rectangular window, spreading energy into neighbouring bins; smoother windows lower the sidelobes.
172. The cross-correlation of two sequences can be computed using FFTs by:
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Correlation is like convolution with one signal reversed.
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Answer: A. Taking the inverse FFT of X[k]·Y*[k], with zero padding to avoid wrap-around
Correlation is convolution with a time-reversed (conjugated) sequence, i.e. multiplication by the conjugate spectrum.
173. Which of the following is NOT a typical application of FFT algorithms?
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Which option is not about discrete spectra or convolution?
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Answer: A. Exact computation of analog filter component values
FFTs are used for spectral analysis, fast convolution/correlation and OFDM; analog component values are found by filter synthesis, not by FFTs.
174. An FIR filter with 64 taps processes audio sampled at 48 kHz in real time. The required number of multiply-accumulate operations per second is:
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Taps × samples per second.
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Answer: C. 3.072 million
Each output needs 64 MACs, so 64 × 48,000 = 3,072,000 MAC/s.
175. A feature found in DSP processors specifically to speed up filtering is:
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Filtering consists of repeated sums of products.
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Answer: D. A single-cycle multiply-accumulate (MAC) unit
Filtering is a sum of products; a hardware MAC with Harvard architecture lets one tap be processed per cycle.
176. In software implementation of an FIR filter on a DSP, the delay line is usually managed with:
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Avoid shifting every sample on each new input.
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Answer: D. A circular buffer using modulo addressing
A circular buffer updates the delay line by moving a pointer instead of shifting all samples; bit-reversed addressing is for FFTs.
177. The filter y[n] = 0.5y[n − 1] + x[n] starts at rest and receives a unit step input. The output y[2] is:
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Iterate the difference equation from n = 0.
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Answer: C. 1.75
y[0] = 1, y[1] = 0.5 + 1 = 1.5, y[2] = 0.75 + 1 = 1.75.
178. For the same filter y[n] = 0.5y[n − 1] + x[n] at rest, the impulse response value h[3] is:
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Iterate with x[0] = 1 and zero input afterwards.
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Answer: B. 0.125
h[n] = (0.5)ⁿu[n], so h[3] = 0.125.
179. The number of complex additions required by an 8-point radix-2 FFT is:
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Count butterflies and additions per butterfly.
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Answer: A. 24
A radix-2 FFT needs N log₂N = 8 × 3 = 24 complex additions (two per butterfly, 12 butterflies).
180. A radix-4 FFT algorithm can be applied directly when the length N is:
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Each stage divides the length by the radix.
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Answer: A. A power of 4
Radix-4 decomposes the DFT into 4-point DFTs at each stage, requiring N = 4^v; mixed or split-radix handles other powers of 2.
181. When implementing a high-order IIR filter in fixed-point hardware, the preferred structure is usually:
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Break the filter into small sections.
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Answer: B. A cascade of second-order sections with proper scaling
Cascaded biquads limit coefficient sensitivity and round-off noise and allow scaling between sections.