Mastering Digital Signal Processing: Your Essential Guide for the FE Electrical Exam
Aug 09, 2024The FE requires examinees for the FE Electrical exam to be able to solve DSP equations. Here, we will provide a brief introduction to the topic and several useful equations that will help prepare you for your upcoming exam.
Sampling
The sampling theorem states that in order to accurately reconstruct a continuous-time signal from its samples, the sampling rate must be at least twice the highest frequency component in the signal. This minimum sampling rate is known as the Nyquist rate. If the sampling rate is below the Nyquist rate, aliasing occurs, leading to distorted or misleading reconstructions.
Digital Filters
Based on the FE Electrical and Computer topic list, the subtopics covered in the Digital Filters are Difference Equation and Z-Transform. However, Finite Impulse Response (FIR) and Infinite Impulse Response (IIR) shall not be neglected even though they were not listed.
1. Difference Equation
A difference equation is a mathematical equation that represents the relationship between the input and output signals of a discrete-time system. It describes how the current and past input values, as well as the current and past output values, are combined to compute the current output value (McClellan et al., 2021, 147).
A general form of a difference equation is:
y[n]=a_{0} x[n]+a_{1} x[n-1]+...+b_{0} y[n-1]+b_{1} y[n-2]+... (x)
Where,
y[n] is the current output at discrete time n.
x[n] is the current input at discrete time n.
x[n-1],... is the past input values.
y[n-1],... is the past output values.
a_{0}, … and b_{0},... are the coefficients of the difference equation.
Example:
y[n]=4x[n]+2y[n-1]+2y[n-2]; y[n]=0 for n<0
What is the impulse response of a system described by the difference equation?
Solution:
1. For impulse response, x[n]=δ[n] and δ[0]=1 and δ[n≥1]=0
. n=0
y[0]=4δ[0]+2y[0-1]+2y[0-2]=4(1)+y[-1]+y[-2]=4+0+0=4
. n=1
y[1]=4δ[1]+2y[1-1]+2y[1-2]=4(0)+2y[0]+2y[-1]=0+2(4)+2(0)=8
. n=2
y[2]=4δ[2]+2y[2-1]+2y[2-2]=4(0)+2y[1]+2y[0]=4+2(8)+2(4)=28
You can keep computing for n approaches ∞ but the solution right is h[n]={4.8,28,...}
2. Z-Transform
The z-transform provides a way to represent discrete-time signals or sequences as functions of a complex variable, denoted as "z." It is defined as the sum of the sequence multiplied by powers of z, where z can be any complex number (McClellan et al., 2021, 348). The z-transform of a discrete-time sequence x[n] is given by:
∞ | |
X(z)=∑ | x[n]z^{-n}(x) |
n = 0 |
In the exam, a table of z transform paris is provided in the Reference Handbook p.372.
Example:
Find the z-transform X(z) of the following signal:
x[n]=δ[n]+1/5 δ[n-2]-3/4 δ[n-4]
Solution:
1. Use the summation definition (equation x) to solve for the z transform:
∞ | |
X(z)=∑ | x[n]z^{-n} |
n = 0 |
2. x[n] is zero for all values of n except when n=0,2,and 4.
x[n]={1,0,1/5,0,-3/4}
. n=0, z transform is
x[0]z^{-0}=1⋅1=1
. n=2, z transform is
x[2]z^{-2}=1/5⋅z^{-2}
. n=4, z transform is
x[4]z^{-2}=-3/4⋅z^{-4}
So, X(z)=1+1/5 z^{-2}-3/4 z^{-4}
3. Finite Impulse Response (FIR)
The FIR system, or, as we often refer to them, FIR filter, is a type of digital filter whose output is determined by a weighted sum of past and present input samples (McClellan et al., 2021, 147).
The output of the FIR filter can be calculated as:
∞ | |
Y(n)=∑ | b_{n}x[n-k] |
k = 0 |
4. Infinite Impulse Response (IIR)
In contrast to FIR filter, the output of IIR filter is the sum of the values of past inputs plus past outputs (McClellan et al., 2021, 394).
The output of the IIR filter can be calculated as:
y[n] = ∑_{l=0}^{∞} b_{l} x[n-l] + ∑_{k=0}^{∞} a_{k} y[n-k]
Example:
With the given filter below
y[n]=x[n]+1/6 x[n-1]
What type of filter is it?
Solution:
1. By looking at the function, the output contains the sum of the current and past inputs. Therefore, it is an FIR filter.
2. Do a z-transform to determine if it is a low-pass, high-pass, band-pass, or band-reject filter.
y[n]=x[n]-1/6 x[n-1]
Use the z-transform pair from Reference Handbook p.372, we have
Y(z)=X(z)-1/6 X(z)z^(-1)
H(z)=(Y(z))/(X(z))=1-1/6 z^(-1)=(z-1/6)/z
Conclusion
This blog aims to serve as a valuable point of reference, offering concise explanations and practical examples for each topic addressed in the exam. It is important to note, however, that while the blog provides a helpful overview, it may not encompass all aspects related to digital signal processing. For a more thorough and comprehensive study plan, I recommend visiting School of PE’s FE Electrical exam review course FE Electrical exam review course, which offers an extensive and detailed curriculum to ensure a comprehensive understanding of the subject matter.
References
D, P., GJ, A., & D, F. (2001). Neuroscience 2nd edition. Sunderland (MA): Sinauer Associates. https://www.ncbi.nlm.nih.gov/books/NBK10924/
Hasegawa, M. (2021). Lecture 6: Sampling and Aliasing. Retrieved June 6, 2023, from https://courses.engr.illinois.edu/ece401/fa2021/lectures/lec06.pdf
McClellan, J. H., Schafer, R. W., & Yoder, M. A. (2021). DSP First Second Edition. Pearson.
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