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What is a non-recursive digital filter?
A non-recursive digital filter calculates the output y[k] using only current and past input values x[k] . It is defined by the difference equation:
y[k] = \sum_{i=0}^{N} a_i x[k - i],
where the coefficients a_i determine the filter's characteristics.
What is the difference equation for a 5-point smoothing filter?
The 5-point smoothing filter has the following difference equation:
y[k] = \frac{1}{35}(-3x[k+2] + 12x[k+1] + 17x[k] + 12x[k-1] - 3x[k-2]).
This is a non-causal filter because the output depends on future inputs like x[k+1] and x[k+2] .
How is causality achieved in a 5-point smoothing filter?
Causality is achieved by introducing a delay, ensuring the output y[k] depends only on current and past inputs. The modified equation is:
y[k] = \frac{1}{35}(-3x[k] + 12x[k-1] + 17x[k-2] + 12x[k-3] - 3x[k-4]).
This delay ensures the filter processes 5 input values before producing the output.
The problem is solved by delaying the calculation of the output value y [k] (the centre point of the parabola) until all the 5 input values have been sampled (i.e. a delay of 2T where T = sampling period).
What does the difference equation represent mathematically?
The difference equation:
y[k] = \sum_{i=0}^N a_i x[k - i]
represents the discrete convolution of the input signal x[k] with the filter coefficients a_i . These coefficients form the impulse response of the filter.
How can the filter coefficients be shown as the impulse response?
To prove the coefficients are the impulse response:
1. Assume x[k] = 0 for all k except x[0] = 1 .
2. Substituting into the difference equation:
y[k] = \sum_{i=0}^N a_i x[k - i].
Since x[k - i] = 0 except when k = i , this simplifies to:
y[k] = a_k.
Therefore, the output y[k] is equal to the filter coefficient a_k , showing that the coefficients form the impulse response.
What is the difference between FIR and IIR filters?
1. FIR (Finite Impulse Response) Filters:
- The impulse response is finite because the number of a_i coefficients is limited.
- Defined as:
y[k] = \sum_{i=0}^N a_i x[k - i].
- Non-recursive filters are always FIR filters.
2. IIR (Infinite Impulse Response) Filters:
- The impulse response is infinite because the output depends recursively on previous outputs:
y[k] = \sum_{i=0}^N a_i x[k - i] + \sum_{i=1}^M b_i y[k - i].
- These must be implemented recursively.
What is the z-transform, and why is it important in digital filtering?
The z-transform describes digital filtering processes in the frequency domain, analogous to how the Laplace transform is used in analogue filtering.
It provides a mathematical framework to analyze and design digital filters, especially regarding sampling, shifting, and convolution.
How is the z-transform defined?
For a discrete-time sequence f[k], the z-transform is:
F(z) = f(0) + f(1)z^{-1} + f(2)z^{-2} + \dots + f(k)z^{-k},
where z = e^{Ts} = e^{\sigma T} e^{j\omega T}.
- z^{-1} implies a time delay of T and can be interpreted as a shift operator.
What are some examples of z-transforms?
1. Step Function: f[k] = 0 for k < 0 and f[k] = 1 for k \geq 1
F(z) = \frac{1}{1 - z^{-1}}.
2. Decaying Exponential: f[k] = e^{-\alpha k T}
F(z) = \frac{1}{1 - e^{-\alpha T}z^{-1}}, \; \text{valid if } |z^{-1}| < 1.
3. Sinusoid: f[k] = \cos(k\omega T)
F(z) = \frac{1 - \cos(\omega T)z^{-1}}{1 - 2\cos(\omega T)z^{-1} + z^{-2}}.
How do you derive the z-transform of a step function?
For the step function:
f[k] = 0 \; \text{for} \; k < 0, \; f[k] = 1 \; \text{for} \; k \geq 1,
the z-transform is:
F(z) = 1 + z^{-1} + z^{-2} + \dots.
This is an infinite geometric series with first term 1 and ratio z^{-1}. The sum of an infinite geometric series is:
F(z) = \frac{1}{1 - z^{-1}}, \quad \text{valid if} \; |z^{-1}| < 1.
How do you derive the z-transform of a decaying exponential?
For the decaying exponential:
f[k] = e^{-\alpha k T},
the z-transform is:
F(z) = 1 + e^{-\alpha T}z^{-1} + e^{-2\alpha T}z^{-2} + \dots.
This is an infinite geometric series with first term 1 and ratio e^{-\alpha T}z^{-1}. The sum is:
F(z) = \frac{1}{1 - e^{-\alpha T}z^{-1}}, \quad \text{valid if} \; |e^{-\alpha T}z^{-1}| < 1.
How do you derive the z-transform of a sinusoid?
For the sinusoid:
f[k] = \cos(k\omega T),
we use Euler's formula:
\cos(k\omega T) = \frac{e^{jk\omega T} + e^{-jk\omega T}}{2}.
The z-transform becomes:
F(z) = \frac{1}{2} \left( \frac{1}{1 - e^{j\omega T}z^{-1}} + \frac{1}{1 - e^{-j\omega T}z^{-1}} \right).
Simplify the sum:
F(z) = \frac{1 - \cos(\omega T)z^{-1}}{1 - 2\cos(\omega T)z^{-1} + z^{-2}}.
What is the pulse transfer function?
The pulse transfer function G(z) is the z-transform equivalent of the transfer function in analogue systems. It is defined as the ratio of the z-transform of the system output Y(z) to the z-transform of the input X(z):
G(z) = \frac{Y(z)}{X(z)}.
For a digital system, the pulse transfer function is the z-transform of the system's impulse response.
How is the pulse transfer function derived?
1. Let the impulse response of the system (e.g., an FIR filter) be:
a_0, a_1, \dots, a_N.
2. The z-transform of the impulse response is:
G(z) = a_0 + a_1z^{-1} + a_2z^{-2} + \dots + a_Nz^{-N}.
3. The input sequence x[k] has the z-transform:
X(z) = x[0] + x[1]z^{-1} + x[2]z^{-2} + \dots.
4. The convolution of a[k] and x[k] produces the output y[k]:
Y(z) = G(z)X(z).
Thus, the pulse transfer function G(z) represents the z-transform of the impulse response.
What is the form of the pulse transfer function for non-recursive filters?
For non-recursive filters, the pulse transfer function is a finite sum:
G(z) = \sum_{i=0}^{N} a_i z^{-i},
where a_i are the filter coefficients.
What is a recursive filter?
A recursive filter is a digital filter where the output depends on both the current and previous inputs as well as previous outputs. It uses feedback to generate the output.
Mathematically, the output in the z-domain takes the following form:
Y(z) = \sum_{i=0}^N a_i z^{-i} X(z) + \sum_{i=1}^M b_i z^{-i} Y(z),
where:
- X(z) is the input signal.
- Y(z) is the output signal.
- a_i are the feedforward coefficients.
- b_i are the feedback coefficients.
The pulse transfer function is then:
G(z) = \frac{Y(z)}{X(z)} = \frac{\sum_{i=0}^N a_i z^{-i}}{1 - \sum_{i=1}^M b_i z^{-i}}.
In a recursive filter:
- The feedback term \sum_{i=1}^M b_i z^{-i} Y(z) makes the output dependent on past outputs.
- Recursive filters are efficient because they require fewer computations but can become unstable if the poles are not inside the unit circle in the z-plane.
What is the pulse transfer function for a recursive filter?
For a recursive digital filter, the pulse transfer function G(z) relates the output Y(z) to the input X(z) as:
G(z) = \frac{\sum_{i=0}^N a_i z^{-i}}{1 - \sum_{i=1}^M b_i z^{-i}},
where:
- a_i are the feedforward coefficients.
- b_i are the feedback coefficients.
How are poles and zeros related to the z-plane in digital filters?
The pulse transfer function G(z) can be represented as a ratio of polynomials, allowing a pole-zero representation:
G(z) = \frac{\prod_{i=1}^N (1 - z/z_{i})}{\prod_{j=1}^M (1 - z/p_{j})},
where:
- z_i are the zeros (roots of the numerator).
- p_j are the poles (roots of the denominator).
The poles and zeros in the z-plane determine the system's frequency response, stability, and causality.
What is the stability criterion for digital filters in the z-plane?
For a digital system to be causal and stable, all poles must lie inside the unit circle in the z-plane. This condition is analogous to the poles being in the left-half of the s-plane for analogue filters.
How is the s-plane mapped to the z-plane?
The mapping from the s-plane to the z-plane is defined by:
z = e^{sT},
where T is the sampling period and s = \sigma + j\omega.
Key mappings include:
1. The imaginary axis of the s-plane (\sigma = 0) maps to the unit circle in the z-plane.
2. The left half-plane of the s-plane (\sigma < 0) maps to the interior of the unit circle.
3. The right half-plane of the s-plane (\sigma > 0) maps to the exterior of the unit circle.
What is the stability condition for this digital filter: y[k] = x[k-1] + \alpha y[k-1]?
1. Compute the z-transform:
Y(z) = X(z)z^{-1} + \alpha Y(z)z^{-1}.
2. Rearrange to find the transfer function:
G(z) = \frac{Y(z)}{X(z)} = \frac{z^{-1}}{1 - \alpha z^{-1}} = \frac{1}{z - \alpha}.
3. The system has a pole at z = \alpha.
4. For stability, the pole must lie inside the unit circle:
|\alpha| < 1.
Since \alpha > 0, the condition becomes:
0 < \alpha < 1.
