A. Introduction

Question 1

Give the relation between cartesian and polar coordinates for a complex number

Answer 1

z = σ + jω = rcos(θ) + jrsin(θ) = r*e^(jθ)

Question 2

What is Euler’s relation?

Answer 2

e^(jz) = cos(z) + jsin(z)
From this (with the complex conjugate) you can also derive expressions for the sine and cosine:
cos(z) = ½( e^(jz) + e^(-jz))
sin(z) = ½*( e^(jz) - e^(-jz))
Note: Euler’s relation can be proven by a Taylor expansion at z=0

Question 3

What is the complex conjugate?

Answer 3

z* = σ - jω = rcos(θ) - jrsin(θ) = r*e^(-jθ)

Question 4

Give the real part, imaginary part, phase, and magnitude of a complex number.

Answer 4

Real part: Re{z} = σ = rcos(θ)
Imaginary part: Im{z} = ω = rsin(θ)
Phase: ∡ z = arctan(ω/σ) = θ
Magnitude: |z|^2 = zz= (σ + jω) (σ – jω) = σ^2 + ω^2
or magnitude: |z|^2 = zz= re^(jθ) * r*e^(-jθ) = r^2

Question 5

What is special about the unit circle?

Answer 5

In the unit circle, we have that r = 1, this gives:
e^(jθ) = cos(θ) + j*sin(θ)
This equation has some very nice properties, which can be seen in the graph.

Question 6

Which coordination system is best suitable for which type of calculations?

Answer 6

Cartesian representation
Use the cartesian representation for addition of complex numbers. Adding complex numbers together is basically vector addition in the complex plane.
z1 = a+jb, z2 = c+jd => z1 + z2 = (a+c) + j(b + d)

Polar representation
Us the polar representation for multiplication of complex numbers. Multiplying complex numbers basically comes down to scaling and rotation in the complex plan.
z1 = r1e^(jθ) , z2 = r2e^(jφ) => z1* z2 = r1* r2* ej^(θ + φ)

Question 7

What is the difference between continuous time (CT) and discrete time (CT) signals?

Answer 7

Continuous time signals can be represented by functions.
x(t) : ℝ → ℝ or ℂ
Discrete time signals only exist at evenly-space, integer time points.
x[n]: ℤ → ℝ or ℂ

Question 8

Give a time-shift transformation for both CT and DT

Answer 8

y(t) = x(t – t0)
y[n] = x[n - n0]

Question 9

Give a reflection (time reversal) transformation for both CT and DT

Answer 9

y(t) = x(-t)
y[n] = x[-n]

Question 10

Give a time expansion transformation for both CT and DT

Answer 10

y(t) = x(t/a), a > 1
y[n] = x(k)[n] = x[n/k] , if n = multiple of k. else: x[n] = 0

Question 11

Give a time contraction transformation for both CT and DT

Answer 11

y(t) = x(at), a > 1

DT signals are not invertible!

Question 12

What does it mean when a signal is periodic?

Answer 12

A signal is periodic when after some time T (CT) or N (DT) it starts to repeat itself.
x(t) = x(t + T), T ∈ ℝ+, ∀t
x[n] = x[n + N], ∀n, N ∈ ℕ+
For example: sine and cosine

Question 13

What are even and odd functions?

Answer 13

Every signal can be decomposed into an even part and an odd part.
Ev{x(t)} = ½(x(t) + x(-t))
Od{x(t)} = ½(x(t) – x(-t))
The signal is then given by: x(t) = Ev{x(t)} + Od{x(t)}

If a signal is even, then Od{x(t)} = 0 and x(t) = x(-t)
If a signal is odd, then Ev{x(t)} = 0 and x(t) = -x(-t)

From the image, we can see that the even part of a signal is mirrored in the y-axis while the odd part of a signal is mirrored in the origin (both x- and y-axis)

Question 14

What is the unit-step function?

Answer 14

A function u(t) (or u[n] for DT) which has the value 1 for all values t>0 (n>=0) and the value 0 for all values t<0 (n<0).

Question 15

What is the Dirac delta?

Answer 15

The Dirac delta is the impulse function for continuous time signals.
As Δ becomes smaller, the height of the rectangle will become closer and closer to infinity

Question 16

What is the unit impulse?

Answer 16

The unit impulse is the impulse function for discrete time signals. It has value 1 for n=0 and the value 0 for all other functions. The unit impulse is also defined as the difference function for the discrete unit-step function.