Signal Processing

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56 Terms

1
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What is the Nyquist sampling theorem?

A signal must be sampled at least twice its highest frequency component to be reconstructed without aliasing.

fs \ge 2f{\max}

This theorem ensures that the sampling rate is sufficient to capture all relevant information in the signal, preventing distortion.

2
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What is aliasing?

When a signal is undersampled, higher frequency components appear as lower frequencies in the sampled signal, distorting the spectrum.

3
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How can aliasing be prevented?

Use an anti-aliasing low-pass filter before sampling to limit the signal bandwidth to below \frac{f_s}{2}.

4
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Define the Nyquist frequency.

f_{N}=\frac{fs}{2} — the highest frequency that can be represented without aliasing at a given sampling rate.

5
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Practical: A signal contains components up to 4\ \mathrm{kHz}. Find the minimum sampling frequency.

Answer:

fs \ge 2f{\max} = 8\ \mathrm{kHz}

6
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Practical: If a signal of 3\ \mathrm{kHz} is sampled at 4\ \mathrm{kHz}, what aliased frequency appears?

Answer:

|fa| = |f - kfs| = |3 - 4| = 1\ \mathrm{kHz}
Steps: Subtract nearest integer multiple of f_s from f to find the folded frequency.

7
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What is quantization in sampling systems?

Rounding continuous-amplitude samples to discrete levels; introduces quantization noise.

8
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Define an ideal low-pass filter in frequency domain.

H(f) = \begin{cases} 1, & |f| \le f_c \\ 0, & |f| > f_c \end{cases}

9
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What is the drawback of an ideal low-pass filter?

It has an infinite impulse response (non-causal, not physically realizable).

10
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What is the difference between analog and digital filters?

Analog filters use resistors, capacitors, and inductors.
Digital filters perform numerical operations on sampled data using difference equations.

11
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List common analog filter types and characteristics.

  • Butterworth: maximally flat magnitude

  • Chebyshev: ripple in passband, steeper roll-off

  • Elliptic: ripple in both pass & stop bands, steepest

  • Bessel: best phase linearity

12
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Define filter order.

The highest power of s (analog) or z (digital) in the denominator of the transfer function.
Higher order → sharper roll-off.

13
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Give the general analog low-pass transfer function.

H(s)=\frac{\omega_{c}^{n}}{s^{n}+a_{n-1}s^{n-1}+\ldots+a_1s+\omega_{c}^{n}} - where 
\omega_{c} is the cutoff frequency n is the filter order.

14
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Define the digital difference equation form of an FIR filter.

y[n] = \sum{k=0}^{M} bk x[n-k]
FIR → finite impulse response, depends only on current and past inputs.

15
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Define the digital difference equation form of an IIR filter.

y[n] = \sum{k=0}^{M} bk x[n-k] - \sum{k=1}^{N} ak y[n-k]
IIR → infinite impulse response, includes feedback from past outputs.

16
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What is the main advantage of FIR filters?

Always stable and can be designed for linear phase response.

17
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What is the main advantage of IIR filters?

Require fewer coefficients (lower order) for a given frequency response; more computationally efficient.

18
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What is the Z-transform of a discrete signal x[n]?

X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}

19
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What is the relationship between the Z-transform and Laplace transform?

z = e^{sT} , where T is the sampling period.
Z-domain is the discrete-time counterpart of the s-domain.

20
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What is the region of convergence (ROC) for the Z-transform?

The set of z values for which the Z-transform sum converges; determines stability and causality.

21
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State the Z-transform time-shift property.

x[n - n0] \;\longleftrightarrow\; z^{-n0} X(z) for a discrete signal x[n]

22
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State the Z-transform convolution property.

x[n] * h[n] \;\longleftrightarrow\; X(z)H(z)

23
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Practical: Find H(z) for the FIR filter y[n] = 0.5x[n] + 0.5x[n-1].

Answer: H(z) = 0.5(1 + z^{-1})
Steps: Take Z-transform of both sides and solve for \frac{Y(z)}{X(z)}.

24
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Practical: Given y[n] = 0.8y[n-1] + x[n], find the transfer function.

Answer: H(z) = \frac{1}{1 - 0.8z^{-1}}
Steps: Take Z-transform: Y(z) - 0.8z^{-1}Y(z) = X(z), solve for H(z) = Y/X.

25
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What is the frequency response of a discrete system?

H(e^{j\omega}) = H(z)\big|_{z = e^{j\omega}}

26
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How is digital filter stability determined?

The system is stable if all poles of H(z) lie inside the unit circle in the z-plane.

27
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Practical: Determine stability of H(z) = \frac{1}{1 - 1.2z^{-1}}.

Answer: Unstable — pole at z = 1.2 (outside unit circle).

28
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What is the sampling period relation to frequency?

T_s = \frac{1}{fs} where fs is the sampling frequency.

29
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What is zero-order hold (ZOH)?

A method of reconstructing analog signals by holding each sample value constant until the next sample.
It approximates the continuous-time signal from discrete samples, maintaining each sample value until the next one is applied.

30
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How do you find the frequency response of a discrete-time system?
Substitute z = e^{j\omega} into the transfer function.
H(e^{j\omega}) = H(z)\big|_{z = e^{j\omega}}
31
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What is the magnitude and phase of a discrete frequency response?

|H(e^{j\omega})| = \sqrt{\text{Re}[H]^2 + \text{Im}[H]^2} \angle H(e^{j\omega}) = \tan^{-1}\!\left(\frac{\text{Im}[H]}{\text{Re}[H]}\right)

32
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Practical: For H(z) = 0.5(1 + z^{-1}), find |H(e^{j\omega})|.

Answer: |H(e^{j\omega})| = |\cos(\omega/2)|

Steps: Substitute z = e^{j\omega} → H = 0.5(1 + e^{-j\omega}) = e^{-j\omega/2}\cos(\omega/2) → magnitude is |\cos(\omega/2)|.

33
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What is the difference between digital and analog angular frequency?
\omega = \Omega T_s
where \omega (rad/sample) is digital frequency, \Omega (rad/s) is analog frequency.
34
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Define normalized digital frequency.
f_{norm} = \frac{f}{f_s}
Ranges from 0 to 0.5 (corresponding to 0 → f_N).
35
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Practical: Convert an analog cutoff frequency of 500\ \mathrm{Hz} to digital if f_s = 2\ \mathrm{kHz}.

Answer: f_{norm} = \frac{500}{2000} = 0.25

36
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Define group delay.

\tau_g(\omega) = -\frac{d\phi(\omega)}{d\omega}

It measures how phase changes with frequency; constant group delay means no phase distortion.

37
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Define linear-phase FIR filter.
A filter where phase response is a straight line:
\phi(\omega) = -\omega n_0
All frequency components are delayed equally.
38
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What are common FIR window design methods?
Rectangular, Hamming, Hanning, Blackman, Kaiser windows.
Tradeoff between main-lobe width (transition sharpness) and side-lobe attenuation.
39
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Which window gives the narrowest main lobe?

Rectangular window — narrowest transition, highest side lobes.

40
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Which window gives the best stopband attenuation?

Blackman or Kaiser window (variable parameter for tradeoff).

41
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What is the Z-plane representation of a digital filter?
Shows poles (X) and zeros (O) in the complex plane.
Poles inside the unit circle → stable system.
42
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How does a zero on the unit circle affect magnitude response?

A zero on the unit circle causes a notch (attenuation) at that frequency.

43
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How does a pole near the unit circle affect magnitude response?
Increases the gain near that frequency (resonant peak).
44
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Practical: A filter has a zero at z = -1. What frequency is attenuated?

Answer: \omega = \pi (Nyquist frequency) **Steps:** z = e^{j\omega} = -1 \Rightarrow \omega = \pi

45
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Practical: A pole at z = 0.9e^{j\pi/4} gives what approximate behavior?

Answer: Resonant amplification near \omega = \pi/4; slowly decaying oscillations in time.

46
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What is the discrete-time impulse response of an FIR filter with coefficients b_0, b_1, b_2?
h[n] = \{b_0, b_1, b_2\} for n = 0,1,2; zero elsewhere.
47
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Define system stability in terms of impulse response.
Stable if the impulse response is **absolutely summable**:
\sum_{n=-\infty}^{\infty} |h[n]| < \infty
48
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What is the effect of downsampling by factor M?
Keeps every M^\text{th} sample, reducing sample rate to f_s/M and compressing spectrum by factor M.
49
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What is the effect of upsampling by factor L?
Inserts L-1 zeros between samples, increasing sample rate to Lf_s and expanding spectrum with image frequencies.
50
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How do you correct distortion after upsampling?

Apply a low-pass reconstruction filter to remove the spectral images.

51
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Practical: A signal sampled at f_s = 10\ \mathrm{kHz} is downsampled by 2. What is new Nyquist frequency?

Answer: f_N = \frac{f_s}{4} = 2.5\ \mathrm{kHz} **Steps:** New sampling rate f_s' = 5\ \mathrm{kHz}, so f_N = f_s'/2 = 2.5\ \mathrm{kHz}.

52
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What is an all-pass filter?
A filter with |H(e^{j\omega})| = 1 for all frequencies but a varying phase.
Used for phase equalization or delay compensation.
53
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What is a notch filter?
A filter designed to reject a narrow band of frequencies; has zeros near the unit circle at the unwanted frequency.
54
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Practical: Design a notch at \omega = \pi/2.

Answer: H(z) = 1 - 2\cos(\pi/2)z^{-1} + z^{-2} = 1 + z^{-2}

55
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Define high-pass and low-pass filters in terms of magnitude response.

Low-pass: passes low frequencies, attenuates high 

High-pass: passes high frequencies, attenuates low

56
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What is the 3 dB cutoff point of a digital filter?
The frequency where
|H(e^{j\omega_c})| = \frac{1}{\sqrt{2}} |H(0)|
#

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