Practice exam

Question 1Not yet answeredMarked out of 3.00Block diagrams

Which of the four block diagrams (a)-(d) (figures in the source) gives an identical result to the reference diagram, where

X(s)

passes through

G(s)

Y(s)

and a branch taps

X(s)

W(s)

(d)(a)(b)(c)

Question 2Not yet answeredMarked out of 3.00Fourier transform

What is the Fourier transform of

x(t) = 8\cos(4t)

X(j\omega) = 4\pi\,(\delta(\omega - 8) + \delta(\omega + 8))

X(j\omega) = 8\pi\,(\delta(\omega - 4) + \delta(\omega + 4))

X(j\omega) = \pi\,(\delta(\omega - 4) + \delta(\omega + 4))

X(j\omega) = 4\left(\frac{\sin(\omega - 8)}{\omega} + \frac{\sin(\omega - 8)}{\omega}\right)

Question 3Not yet answeredMarked out of 3.00Ideal op-amp

What is not a correct assumption for the ideal operational amplifier?

The differential voltage between the two inputs is 0.The gain of the input signal is 0.The current at the positive input is 0.The current at the negative input is 0.

Question 4Not yet answeredMarked out of 3.00Signal classification

A continuous signal

x(t)

(figure in the source) starts flat at zero, rises into a smooth hump after

t = 0

, and settles back down. Which property does not apply to it?

DeterministicCausalAnalogOdd

Question 5Not yet answeredMarked out of 3.00Impulse response

A system has the step response

h(t) = \cos(5t) + 3\sin(2t)

. What is the weighting function (impulse response) of the system?

g(t) = 6\cos(5t) - 2\sin(5t)

g(t) = 3\cos(2t) + \sin(5t)

g(t) = 6\cos(2t) - 5\sin(5t)

g(t) = \cos(2t) - 5\sin(5t)

Question 6Not yet answeredMarked out of 3.00Signal spaces

Which of the following signals does not belong to a common signal space of the other three?

A_1 \sin(2\pi f k + \varphi_1)

A_1 \cos(2\pi f k + \varphi_1)

A_2 \sin(2\pi f k + \varphi_2)

A_1 \sin(2\pi (f + c) k + \varphi_2),\ c \neq 0

Question 7Not yet answeredMarked out of 3.00Signal energy

What is the signal energy of

x(t) = 3 + 4j

for

|t| \le 2

and

0

otherwise?

E_x = 500

E_x = 50

E_x = 100

E_x = 25

Question 8Not yet answeredMarked out of 6.00Sampling

(a) What sampling interval step size does a sampling frequency of

f_a = 500\ \mathrm{Hz}

correspond to? (b) By multiplication with which function is ideal sampling described? Give it for this frequency. (c) A real sampling has conversion time

T_\omega

. What does this do to the frequency spectrum (qualitatively)? (d) Calculate the first sampled point over

t = 0 \dots 0.3\ \mathrm{s}

for

x(t) = 5\cos(t)

with conversion time

T_\omega = 0.3\ \mathrm{s}

Question 9Not yet answeredMarked out of 8.00Fourier series and transform

(a) For which assumption does the Fourier series become the Fourier transform? (b) Compute the Fourier transform of

x(t) = 5

for

-7 \le t \le 7

and

0

otherwise, distinguishing

\omega \neq 0

and

\omega = 0

. (c) The signal is periodically continued with period

T_0 = 20

. Use the spectrum to find the Fourier coefficients

c_2

and

c_4

Question 10Not yet answeredMarked out of 10.00Discrete-time transforms

(a) Calculate the discrete-time Fourier transform of

x_1[k] = 0.7^k\,\sigma[k]

and

x_2[k] = 0.3^k\cos(6k)\,\sigma[k]

. (b) For the sampled signal

x[k] = 3, 5, 0

k = 0, 1, 2

, calculate the first two DFT coefficients

X[n]

n = 0, 1

(real and imaginary parts). (c) How can the spectrum from (b) be improved to have 12 points per period?

You are allowed to use that.

Unit 1Signals and Systems

An infinitely thin, infinitely tall spike of area 1.

The sifting property: multiplying by $\delta(t-t_0)$ and integrating just picks out $x(t_0)$ .

Any signal splits into an even part (mirror-symmetric) and an odd part (point-symmetric).

A signal is odd only if $x(-t) = -x(t)$ ; a signal starting flat then rising is not.

Seen in Practice exam 3489.

Use $|x|^2$ , not $x^2$ : for a complex value $3 + 4j$ it is $|3+4j|^2 = 25$ .

Seen in Practice exam 3489.

Energy signals have finite energy; power signals (like a never-ending sine) have finite power.

$A$ is amplitude, $\omega$ angular frequency, $\varphi$ phase.

A switch that turns on at $t = 0$ and stays on. Also written $u(t)$ .

Unit 2Time-Domain Analysis (LTI Systems)

Bounded input gives bounded output exactly when the impulse response is absolutely integrable.

The output of any LTI system is its input convolved with the impulse response $h$ .

Flip $h$ , slide it across $x$ , multiply, and add up the overlap at each shift. See the convolution note.

Poke the system with an impulse and $h(t)$ is what comes out.

Once you know $h$ , you know the response to everything (via convolution).

Zero-input response (from initial conditions) plus zero-state response (from the input).

The weighting function (impulse response) is the derivative of the step response $h_\sigma$ .

Because the impulse is the derivative of the step, their responses are too.

Seen in Practice exam 3489.

Unit 3The Laplace Transform

A single forward block with a branch tapped before it is reproduced by a parallel $1/G$ on the branch.

Seen in Practice exam 3489.

Inverse transforms are done by matching a table after partial fractions.

The whole point of transforms: messy convolution in time is just a product in $s$ .

Each derivative brings down a factor of $s$ ; this is how ODEs become polynomials.

Infinite open-loop gain forces the input difference to zero; infinite input impedance forces the input currents to zero.

The *closed-loop* gain is set by the feedback, and is certainly not zero.

Seen in Practice exam 3489.

Turns differential equations into algebra by trading $t$ for the complex frequency $s = \sigma + j\omega$ .

The Laplace transform of the impulse response; output over input in the $s$ -domain.

Unit 4Fourier Series and Transform

$N$ samples give $N$ spectrum points. Zero-padding to more samples gives a finer spectrum.

Seen in Practice exam 3489.

For a decaying geometric $a^k\sigma[k]$ it sums to $\dfrac{1}{1 - a\,e^{-j\Omega}}$ (a geometric series).

Seen in Practice exam 3489.

A periodic signal is a sum of harmonics; $c_k$ is the amount of the $k$ -th one.

Seen in Practice exam 3489.

The Fourier series taken to the limit as the period $T_0 \to \infty$ ; the spectrum becomes continuous.

Seen in Practice exam 3489.

Ideal sampling multiplies the signal by this comb of impulses.

In frequency this makes the spectrum repeat periodically; overlap is aliasing.

Seen in Practice exam 3489.

Rebuild the time signal by adding its frequency components back up.

Energy is the same whether you measure it in time or in frequency.

The time between samples is the reciprocal of the sampling frequency.

Seen in Practice exam 3489.

A pure tone is two impulses in frequency, at $\pm\omega_0$ . A sine is the same but imaginary and antisymmetric.

Seen in Practice exam 3489.

A box in time is a sinc in frequency. At $\omega = 0$ take the limit: $X(0) = A\tau$ (the area).

Seen in Practice exam 3489.

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Practice exam

Practice exam

Formulary

Unit 1Signals and Systems

Unit 2Time-Domain Analysis (LTI Systems)

Unit 3The Laplace Transform

Unit 4Fourier Series and Transform

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