Problem #1: Classification of Signal and Systems

Consider the signal and system shown below. The system is characterized by an input/output relation. Think of it as follows: when you apply a voltage for the first time, the output starts at zero volts. As you increase the voltage, the output increases as shown on part A of the curve until you reach a maximum. As you subsequently decrease the voltage, the output drops along curve B. As you continue to decrease the voltage, the output varies along C, etc. As you continue to cycle the voltage, the output never returns to curve A, but remains in the cycle B,C,D. You might consider this a simple model of a magnet.

(a) Classify the system using as many concepts discussed in Chapters 1-3 as possible. Be sure to

justify your answer ---answers with no justification get no credit.

The system is dynamic because the output depends on past or future values in addition to the present.

The system is causal because the output does not anticipate the input.

The system in non-linear because superposition does not hold.

The system is continous in time because it is continous at every point in time.

(b) Represent the input signal, x(t) as a combination of singularity functions.

The signal can be seperated into three parts as seen below:

(1) Take the slope of the ramp function and multiply it by the impulse function.

(2) This is the same as part 1 except it is a negative slope with a shift in time.

(3) The third part is an impulse function with a shift in time as seen below.

If all three steps are added, the results will be as follows:

x(t) = 2r(t) (2t - 1/2) - 2r(t - 1/2) (2t - 3/2) + (t - 3/2)

(C) Compute the power and the energy of the input signal.

Since there is a finite value, the system is an Energy Signal.

If the system is an Energy Signal, the Power Signal is zero or

The integration will be the same for the power signal calculations with the exception of (1/2T) where T approaches infinity making the power signal equal to zero.

(d) Compute the DC value of the input signal.

The DC value of the input signal is zero because x(t) is aperiodic, a0 = 0.

Problem #2: Time-Domain Solutions

Consider the signal and system (described by its implulse response) shown below:

(a) Compute the output, y(t), for the system shown below:

To find y(t), x(t) is convolved with h(t): y(t) = x(t) * h(t).

After the convolution has been done, the resulting graph, y(t), will be as follows:

(b) Is the system causal? Explain?

No, the system is non-causal because h(t) starts before t=0, therefore the system depends upon future values of the input.

Problem #3: Fourier Series

One period of a periodic signal is shown below:

(a) Using symmetry arguments, explain which Fourier coefficients of the trigonometric Fourier

series will be zero ({an} and {bn}).

The waveform shown above is periodic, which is defined for all time t, where t( -∞,∞). This means the exact same waveform will appear on the left-hand side as follows.

This shows that f(t) = f(-t), which proves the case of even symmetry. Also, the waveform shows that f(t) = -f(t-To/2) centered around the DC value of the signal, which is known as half-wave symmetry. Putting these cases together, the waveform is said to be half-wave even symmetry.

The coefficients of the waveform will be as follows:

where

bn = 0, for all n

an = bn = 0, for n odd

(b) Compute the first term in the trigonometric Fourier series, ao.

(C) Compute the remaining coefficients of the trigonometric Fourier series.

The remaining coeficients are an with n being odd.

Therefore,

(d) Compute the power of the output signal, y(t).

To compute the power, Parseval's theorem for the frequency domain is used:

where Yn is the magnitude of the Fourier coeficients and Y0 is the average DC value.

Since the linear system removes all frequencies above 30 Hz, the output will only contain the dc value and the fundamental harmonic, Yo and Y1 respectively. Therefore, n = 1 for the following solutions:

Therefore,