Linear System Signal Analysis Practice

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1.1Classify a Signal

Exam I | Problem 1.1 |

Classify the signal below as continuous-time or discrete-time, and state whether it is deterministic or random:

x(t) = e^{-2t}u(t)

Solution

The signal is continuous-time because it is defined for real values of $t$.

It is deterministic because it is given by an explicit formula, not a statistical description.

1.2Shift a Step Signal

Exam I | Problem 1.2 |

Time Shifting

$$ x(t) = u(t), $$

what is $x(t-3)$?

Solution

Replace $t$ by $t-3$:

$$ x(t-3) = u(t-3) $$

This is the original step delayed by $3$ units.

1.3Find Even and Odd Parts

Exam I | Problem 1.3 |

Even and Odd Parts

Let

$$ x(t) = t + 2. $$

Find the even part $x_e(t)$ and odd part $x_o(t)$.

Solution

First compute $x(-t)$:

$$ x(-t) = -t + 2 $$

Now use the formulas:

x_e(t) = \frac{x(t) + x(-t)}{2}

x_o(t) = \frac{x(t) - x(-t)}{2}

x_e(t) = \frac{(t+2)+(-t+2)}{2} = 2

and

x_o(t) = \frac{(t+2)-(-t+2)}{2} = t

1.4Test Linearity

Exam I | Problem 1.4 |

Linearity

Is the system

$$ y(t) = x(t)^2 $$

linear?

Solution

No. Linearity requires superposition.

For an input $x(t)$, the output is $x(t)^2$. But if we double the input, the output becomes

$$ (2x(t))^2 = 4x(t)^2 $$

whereas doubling the original output gives

$$ 2x(t)^2 $$

These are not the same, so the system is not linear.

1.5Check Memory and Causality

Exam I | Problem 1.5 |

For the system

$$ y(t) = x(t) + x(t-1), $$

state whether the system has memory and whether it is causal.

Solution

The system has memory because $y(t)$ depends on $x(t-1)$, not just the input at the same time.

It is causal because it depends only on the present value $x(t)$ and the past value $x(t-1)$, not on any future input.

1.6Use the Impulse Shift Property

Exam I | Problem 1.6 |

For an LTI system with impulse response $h(t)$, what is the output when the input is

x(t) = \delta(t-2)?

Solution

An impulse delayed by $2$ units produces a delayed copy of the impulse response:

$$ y(t) = h(t-2) $$

This follows from the shift property of convolution.

1.7Convolve with a Delayed Impulse

Exam I | Problem 1.7 |

Given

h[n] = \delta[n] + \delta[n-1]

and

x[n] = \delta[n-2],

find $y[n] = x[n] * h[n]$.

Solution

Convolution with a delayed impulse shifts the other signal:

$$ y[n] = h[n-2] $$

y[n] = \delta[n-2] + \delta[n-3]

1.8Find the Fundamental Angular Frequency

Exam I | Problem 1.8 |

A periodic signal has period

T = 0.25 \text{ s}.

What is its fundamental angular frequency $\omega_0$?

Solution

Use

\omega_0 = \frac{2\pi}{T}

Substitute $T = 0.25$:

\omega_0 = \frac{2\pi}{0.25} = 8\pi

So the fundamental angular frequency is

8\pi \text{ rad/s}

1.9Read a Pole from a Transfer Function

Exam I | Problem 1.9 |

For

H(s) = \frac{1}{s+5},

find the pole and state whether it lies in the left-half plane or right-half plane.

Solution

The pole is where the denominator is zero:

s + 5 = 0 \quad \Rightarrow \quad s = -5

Since $-5$ has negative real part, the pole lies in the left-half plane.

1.10Read a Scalar State-Space Model

Exam I | Problem 1.10 |

For the system

\dot{x}(t) = -3x(t) + 2u(t), \qquad y(t) = 4x(t) - u(t),

identify $A$, $B$, $C$, and $D$.

Solution

Compare the equations to

\dot{x}(t) = Ax(t) + Bu(t)

and

$$ y(t) = Cx(t) + Du(t). $$

A = -3,\quad B = 2,\quad C = 4,\quad D = -1.

Difficulty

Integrated

2.1Classify a Differential-Equation Model

Exam II | Problem 2.1 |

Determine whether the system

\frac{dy(t)}{dt} + 2y(t) = \frac{dx(t)}{dt} + 3x(t)

is linear and time invariant.

Solution

The system is linear because $x(t)$ and $y(t)$ appear only to the first power and are not multiplied together.

It is time invariant because the coefficients are constant and there is no explicit dependence on $t$.

So the system is LTI.

2.2Split a Polynomial Signal

Exam II | Problem 2.2 |

Even and Odd Parts

Let

$$ x(t) = t^2 + 3t. $$

Find the even part and odd part of the signal.

Solution

First compute

$$ x(-t) = t^2 - 3t $$

Then

x_e(t) = \frac{x(t) + x(-t)}{2}

and

x_o(t) = \frac{x(t) - x(-t)}{2}

x_e(t) = \frac{(t^2+3t)+(t^2-3t)}{2} = t^2

and

x_o(t) = \frac{(t^2+3t)-(t^2-3t)}{2} = 3t

2.3Output of a Moving-Average Filter

Exam II | Problem 2.3 |

Let

h[n] = \frac{1}{3}\big(\delta[n] + \delta[n-1] + \delta[n-2]\big)

and

$$ x[n] = u[n] - u[n-3]. $$

Find $y[n] = x[n] * h[n]$.

Solution

The filter averages three consecutive samples:

y[n] = \frac{1}{3}\big(x[n] + x[n-1] + x[n-2]\big)

Since $x[n]$ is a length-3 pulse, the output is

y[n] = \begin{cases} \frac{1}{3}, & n=0 \\ \frac{2}{3}, & n=1 \\ 1, & n=2 \\ \frac{2}{3}, & n=3 \\ \frac{1}{3}, & n=4 \\ 0, & \text{otherwise} \end{cases}

2.4Compute a Sinusoidal Steady-State Output

Exam II | Problem 2.4 |

An LTI system has frequency response

H(j\omega) = 2e^{-j\pi/6}

at $\omega = 5$. If the input is

x(t) = \cos(5t),

what is the steady-state output?

Solution

For a cosine input at frequency $\omega$, the output keeps the same frequency and is scaled and phase shifted by $H(j\omega)$.

The magnitude is $2$ and the phase is $-\pi/6$, so

y(t) = 2\cos\left(5t - \frac{\pi}{6}\right)

2.5Shift a Known Fourier Transform

Exam II | Problem 2.5 |

Suppose

x(t) = e^{-at}u(t), \qquad a > 0,

has Fourier transform

X(j\omega) = \frac{1}{a + j\omega}.

What is the Fourier transform of $x(t-3)$?

Solution

Use the time-shift property:

x(t-t_0) \longleftrightarrow e^{-j\omega t_0}X(j\omega)

With $t_0 = 3$,

\mathcal{F}\{x(t-3)\} = e^{-j3\omega}X(j\omega)

\mathcal{F}\{x(t-3)\} = \frac{e^{-j3\omega}}{a + j\omega}

2.6Solve a First-Order System with Laplace Transforms

Exam II | Problem 2.6 |

Solve for $y(t)$ given

y'(t) + 4y(t) = u(t), \qquad y(0)=0.

Solution

Take the Laplace transform of both sides:

sY(s) + 4Y(s) = \frac{1}{s}

Y(s) = \frac{1}{s(s+4)}

Use partial fractions:

\frac{1}{s(s+4)} = \frac{1}{4}\left(\frac{1}{s} - \frac{1}{s+4}\right)

Invert term by term:

y(t) = \frac{1}{4}\left(1 - e^{-4t}\right)u(t)

2.7Find the Transfer Function of a Difference Equation

Exam II | Problem 2.7 |

For the causal system

$$ y[n] - 0.6y[n-1] = x[n], $$

find the transfer function $H(z)$ and the ROC.

Solution

Take the Z-transform with zero initial conditions:

Y(z) - 0.6z^{-1}Y(z) = X(z)

Factor out $Y(z)$:

Y(z)\big(1 - 0.6z^{-1}\big) = X(z)

H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1 - 0.6z^{-1}}

For the causal system, the ROC is

$$ |z| > 0.6 $$

2.8Apply the Nyquist Criterion

Exam II | Problem 2.8 |

A signal has highest frequency content at $900$ Hz. What is the Nyquist rate, and is sampling at $1.5$ kHz sufficient to avoid aliasing?

Solution

The Nyquist rate is

2f_{\max} = 2(900) = 1800 \text{ Hz}

So the sampling rate must be greater than $1800$ Hz to avoid aliasing.

Since $1.5$ kHz $= 1500$ Hz is below $1800$ Hz, it is not sufficient.

Difficulty

Applied

3.1Impulse Input as a Weighted Sum

Final | Problem 3.1 |

An LTI system has impulse response

h(t) = e^{-2t}u(t).

If the input is

x(t) = 3\delta(t-1) - 2\delta(t) + \delta(t-3),

find the output $y(t)$.

Solution

Use linearity and the shift property:

$$ y(t) = 3h(t-1) - 2h(t) + h(t-3) $$

Substitute $h(t)$:

y(t) = 3e^{-2(t-1)}u(t-1) - 2e^{-2t}u(t) + e^{-2(t-3)}u(t-3)

3.2Attenuation of a Sinusoid

Final | Problem 3.2 |

An LTI system has frequency response

H(j\omega) = \frac{1}{1+j\omega}.

If the input is

x(t) = 5\cos(2t),

find the steady-state output amplitude and phase shift.

Solution

Evaluate the frequency response at $\omega=2$:

H(j2) = \frac{1}{1+j2}

Its magnitude is

|H(j2)| = \frac{1}{\sqrt{1^2+2^2}} = \frac{1}{\sqrt{5}}

and its phase is

\angle H(j2) = -\tan^{-1}(2)

So the output amplitude is

5 \cdot \frac{1}{\sqrt{5}} = \sqrt{5}

and the steady-state output is

y(t) = \sqrt{5}\cos\big(2t - \tan^{-1}(2)\big)

3.3Aliased Tone

Final | Problem 3.3 |

A $3.4$ kHz sinusoid is sampled at $4$ kHz. What aliased frequency appears after sampling?

Solution

The sampled spectrum repeats every $f_s = 4$ kHz.

Since

3.4 \text{ kHz} = 4.0 \text{ kHz} - 0.6 \text{ kHz},

the tone folds to

0.6 \text{ kHz} = 600 \text{ Hz}

in the baseband interval $[0, f_s/2]$.

3.4State-Space to Transfer Function

Final | Problem 3.4 |

For the scalar state-space model

\dot{x}(t) = -2x(t) + u(t), \qquad y(t) = 3x(t) + 4u(t),

find the transfer function $H(s)$.

Solution

Use

H(s) = C(sI-A)^{-1}B + D

with

A=-2,\quad B=1,\quad C=3,\quad D=4.

Then

H(s) = 3(s+2)^{-1}(1) + 4 = \frac{3}{s+2} + 4

H(s) = \frac{4s+11}{s+2}

3.5Interpret a Symmetric Periodic Waveform

Final | Problem 3.5 |

A real periodic signal has period

T = 0.01 \text{ s}

and odd symmetry:

$$ x(-t) = -x(t). $$

What kinds of Fourier-series terms can appear, and what is the fundamental frequency?

Solution

Odd symmetry means the Fourier series can contain sine terms only. There is no DC term and no cosine terms.

The fundamental frequency is

f_0 = \frac{1}{T} = \frac{1}{0.01} = 100 \text{ Hz}

So the harmonics occur at integer multiples of $100$ Hz.

Difficulty

Challenge

4.1Convolution of Two Rectangular Pulses

Final | Problem 4.1 |

Let

$$ x(t) = u(t) - u(t-2) $$

and

$$ h(t) = u(t) - u(t-1). $$

Find

$$ y(t) = x(t) * h(t). $$

Solution

Both signals are rectangular pulses. The convolution equals the amount of overlap between the interval $[0,2)$ and the shifted interval $[t-1,t)$.

That overlap length is:

y(t) = \begin{cases} 0, & t < 0 \\ t, & 0 \le t < 1 \\ 1, & 1 \le t < 2 \\ 3-t, & 2 \le t < 3 \\ 0, & t \ge 3 \end{cases}

So the output is a trapezoid made of two ramps and a flat middle section.

4.2Solve a Forced First-Order System

Final | Problem 4.2 |

Solve for $y(t)$ given

y'(t) + 3y(t) = e^{-2t}u(t), \qquad y(0)=0.

Solution

Take the Laplace transform:

sY(s) + 3Y(s) = \frac{1}{s+2}

Y(s) = \frac{1}{(s+2)(s+3)}

Use partial fractions:

\frac{1}{(s+2)(s+3)} = \frac{1}{s+2} - \frac{1}{s+3}

Invert term by term:

y(t) = \big(e^{-2t} - e^{-3t}\big)u(t)

4.3Stability and Causality from Poles

Final | Problem 4.3 |

Consider

H(s) = \frac{s+1}{(s+2)(s-3)}.

Can a causal realization of this system be stable? Explain.

Solution

The poles are at

s = -2 \quad \text{and} \quad s = 3.

For a causal continuous-time rational system, the ROC must lie to the right of the rightmost pole, so here the ROC would be

\Re(s) > 3.

That ROC does not include the $j\omega$ axis, so the system would not be stable.

Therefore, a causal stable realization is not possible.

4.4Frequency Response from an Exponential Impulse Response

Final | Problem 4.4 |

A system has impulse response

h(t) = e^{-at}u(t), \qquad a > 0.

For the input

x(t) = \cos(\omega t),

find the steady-state output amplitude and phase shift.

Solution

The frequency response is

H(j\omega) = \frac{1}{a+j\omega}

so its magnitude is

|H(j\omega)| = \frac{1}{\sqrt{a^2+\omega^2}}

and its phase is

\angle H(j\omega) = -\tan^{-1}\left(\frac{\omega}{a}\right)

Therefore the steady-state output is

y(t) = \frac{1}{\sqrt{a^2+\omega^2}}\cos\left(\omega t - \tan^{-1}\left(\frac{\omega}{a}\right)\right)