Name: Stephen C. Schruff

ProblemPointsScore

1a10
1b10
2a10
2b10
2c10
3a10
3b10
3c10
3d10
3e10
Total100

Notes:

1.	The exam is closed books/closed notes - except for one page of notes.

2.	Please show ALL work. Incorrect answers with no supporting explanations or work will be given no partial credit.

3.	Please indicate clearly your answer to the problem. If I can't read it (and I am the judge of legibility), it is wrong. If I can't follow your solution (and I get lost easily), it is wrong. All things being equal, neat and legible work will get the higher grade:)

Problem No. 1: Show the following properties of the Fourier transform for real signals:

(a) |X(f)| = |X(-f)|

X(f) may be written in terms of magnitude and phase as follows:

Since,

(b) (f) = - (f)

Using A( ) and B( ) from the above proof, it can be shown that:

Therefore,

Problem No. 2: For the following two signals:

(a)	Plot the magnitude spectrum of x(t):

(b)	Plot the magnitude spectrum of y(t):

( c)	Compare and contrast these plots (note that you can answer this question correctly without getting the correct answers for (a) and (b).

x(t) is a unit pulse function that is shifted to the right by ½ second.

y(t) is a unit pulse function that is expanded from a width of one second to two seconds.

y(t) has a magnitude twice as large as x(t) because it is twice as wide in the time domain as x(t).

Therefore, the gain of y(t) is twice as large as that of x(t).

The frequency of y(t) is half that of x(t) because y(t) is twice as large as x(t) in the time domain.

The two previous descriptions are due to the Fourier Scale Change Theorem. It is stated as

follows:

The two plots shown are the magnitudes of the signals. It is important to note that the magnitude

of x(t) is not affected by the shift in time. This is proven by the Fourier Time Delay Theorem.

However, the shift in time does affect the phase spectrum of the signal.

Problem No. 3: For the following system,

(a) Find X(s) :

Find x(t) using the equation of a line:

(b) Find Y(s):

where

( c) For the circuit shown, construct the Bode magnitude plot.

|H(s)| = 0 at = 0 because 20 log ( 1 ) = 0

There is a double zero at = 0; therefore, the magnitude increases at a rate of 40 dB per decade.

There is a double pole at = 1; therefore, the magnitude remains constant.

(d) How many poles and zeros does the system have?

It has a double pole at = 1. Thus, it has 2 poles.

It has a double zero at = 0. Thus, it has 2 zeroes.

(e) Using concepts introduced in Chaps. 4 - 6, determine the stability of the system.

The system is marginally stable because it has no poles in the right half plane and no poles in

the left half plane. The poles are on the imaginary axis at -j and +j. This makes it a marginally

stable system.This can be proven with the Routh Array. However, this is not necessary since this

is a simple LC circuit with capacitor and inductor values that are equal.

A stable system cannot have unbounded magnitude. The impedance of the inductor is sL and

that of the capacitor is 1/sC. Therefore, the two elements of the system counterbalance

each other making for a marginally stable system.