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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.
What is aliasing?
When a signal is undersampled, higher frequency components appear as lower frequencies in the sampled signal, distorting the spectrum.
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}.
Define the Nyquist frequency.
f_{N}=\frac{fs}{2} — the highest frequency that can be represented without aliasing at a given sampling rate.
Practical: A signal contains components up to 4\ \mathrm{kHz}. Find the minimum sampling frequency.
Answer:
fs \ge 2f{\max} = 8\ \mathrm{kHz}
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.
What is quantization in sampling systems?
Rounding continuous-amplitude samples to discrete levels; introduces quantization noise.
Define an ideal low-pass filter in frequency domain.
H(f) = \begin{cases} 1, & |f| \le f_c \\ 0, & |f| > f_c \end{cases}
What is the drawback of an ideal low-pass filter?
It has an infinite impulse response (non-causal, not physically realizable).
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.
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
Define filter order.
The highest power of s (analog) or z (digital) in the denominator of the transfer function.
Higher order → sharper roll-off.
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.
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.
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.
What is the main advantage of FIR filters?
Always stable and can be designed for linear phase response.
What is the main advantage of IIR filters?
Require fewer coefficients (lower order) for a given frequency response; more computationally efficient.
What is the Z-transform of a discrete signal x[n]?
X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}
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.
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.
State the Z-transform time-shift property.
x[n - n0] \;\longleftrightarrow\; z^{-n0} X(z) for a discrete signal x[n]
State the Z-transform convolution property.
x[n] * h[n] \;\longleftrightarrow\; X(z)H(z)
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)}.
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.
What is the frequency response of a discrete system?
H(e^{j\omega}) = H(z)\big|_{z = e^{j\omega}}
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.
Practical: Determine stability of H(z) = \frac{1}{1 - 1.2z^{-1}}.
Answer: Unstable — pole at z = 1.2 (outside unit circle).
What is the sampling period relation to frequency?
T_s = \frac{1}{fs} where fs is the sampling frequency.
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.
|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)
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)|.
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
\tau_g(\omega) = -\frac{d\phi(\omega)}{d\omega}
It measures how phase changes with frequency; constant group delay means no phase distortion.
Rectangular window — narrowest transition, highest side lobes.
Blackman or Kaiser window (variable parameter for tradeoff).
A zero on the unit circle causes a notch (attenuation) at that frequency.
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
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.
Apply a low-pass reconstruction filter to remove the spectral images.
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}.
Practical: Design a notch at \omega = \pi/2.
Answer: H(z) = 1 - 2\cos(\pi/2)z^{-1} + z^{-2} = 1 + z^{-2}
Low-pass: passes low frequencies, attenuates high
High-pass: passes high frequencies, attenuates low