Midterm #1 for the Fall 2017 semester will be on Thursday, Oct. 5th, during lecture time (12:30pm to 2:00pm) but in the following rooms:
You will also be responsible for the material in
There will likely be five questions on Midterm #1. There will be no questions about Matlab commands or syntax on the midterm.
The Web site that accompanies the Signal Processing First book has dozens of worked homework problems from chapters 2-4 of DSP First. Chapters 2-4 of DSP First cover identical material as chapters 2-4 of Signal Processing First. You can access the homework problems and their solutions via the Homework menu at the top of the Web page, or by using the Homework link in the supplemental material for each chapter.
Here are the questions from the previous exams that are related to the material to be covered on midterm #1 this semester:
Problem 2: You can solve this problem with using only one addition (1+1) and one multiplication (1 times 1).
(a) Ideas used in finding the answer without doing any calculations:
(b) Ideas used to find the answer without any calculations:
Problem 3: This problem involves very little math. It is meant to test concepts. (As a side note: if the description of a system response involves a summation, it does not necessarily mean that the system is a discrete-time system. Conversely, if a description of a system response involves an integral, it does not necessarily mean that the system is a continuous-time system.)
(a) Finite impulse response
(b) The impulse response is the system response when an impulse is input. Since the system is continuous time, use a Dirac delta functional d(t) for the impulse: x(t) = d(t). So, in the summation for y(t), replace x with d.
(c) The step response is the system response when a step function is input. So, in the summation for y(t), replace x with u.
(d) For N = 3, the step response is
(e) (N - 1) T
Problem 4: The solution to part (b) gets to very tedious. In the future, I would try to not assign a problem this tedious on a midterm.
(a) The characteristic equation is 1 - 3/2 D-1 + K D-2 = 0. So, there are two roots:
(b) The roots need to be inside the unit circle. So, | r0 | < 1 and | r1 | < 1. Solve for K. This is the tedious part. The answer is something like 1/2 < K < 1.
It means that if there are non-zero initial conditions, the system will output a weighted combination of its characteristic modes. That output would be sustained for all time from time 0 to time infinity.
Question 2: I get confused about what types of systems can be used for what types of filters, and i can't really find a specific section in the book about it, can you direct me toward where I might find a better understanding of what systems make what filters and what applications they can have?
This notion will be more clear after we learn more about frequency responses (Laplace and Fourier transforms) in the second part of the course.
As far as midterm #1, we have seen two examples of a lowpass filter (integrator in continuous-time and an averager in discrete-time) and a highpass filter (differentiator in continuous-time and a first-order difference in discrete-time). We have also seen one example of an all-pass filter (homework problem 3.3).
The Mandrill (Baboon) demonstration uses a cascade of a lowpass and a highpass filter, and the cascade has a bandpass response.
I have not presented any examples of bandstop filters yet.
Question 3: Finally, in my differential equations class, we didn't really use the ej t for complex roots, we just made it as sin t + cos t. I'm used to solving it this way, so the book method of using ej t or cos(t + theta) is kind of confusing. Should I use my time to learn this method or would the other way be ok on the test?
In terms of solving a differential equation, what matters is getting the right answer with a mathematically correct method.
In terms of understanding the behavior of systems governed by differential equations, it is important to know how the roots of the characteristic polynomial are mapped into characteristic modes. This is where the elambda t and tk elambda t forms arise.