Signals
Recall (Signal) A signal is a set of information or data. Any physical quantity that varies over time, space or any other variable or variables.
Even, Odd & Periodic
An even function is symmetrical about the vertical axis. Mathematically this means:
f ( t ) = f ( − t ) f [ k ] = f [ − k ] f(t) = f(-t) \newline
f[k] = f[-k] f ( t ) = f ( − t ) f [ k ] = f [ − k ] An odd function is anti-symmetrical about the vertical axis. Mathematically this means:
f ( t ) = − f ( − t ) f [ k ] = − f [ − k ] f(t) = -f(-t) \newline
f[k] = -f[-k] f ( t ) = − f ( − t ) f [ k ] = − f [ − k ] A periodic function has a fundamental period (minimum), T 0 T_0 T 0 fundamental frequency , f 0 = 1 T 0 f_0 = \dfrac{1}{T_0} f 0 = T 0 1
f ( t ) = f ( t + n T 0 ) f [ k ] = f [ k + n K 0 ] f(t) = f(t + nT_0) \newline
f[k] = f[k + nK_0] f ( t ) = f ( t + n T 0 ) f [ k ] = f [ k + n K 0 ] We sometimes define the fundamental frequency in angular velocity instead of Hz, which means, ω 0 = 2 π f 0 = 2 π T 0 \omega_0 = 2\pi f_0 = \dfrac{2\pi}{T_0} ω 0 = 2 π f 0 = T 0 2 π
Convolution
Let’s call our input signal x ( t ) x(t) x ( t ) h ( t ) h(t) h ( t ) y ( t ) y(t) y ( t ) ∗ * ∗
y ( t ) = x ( t ) ∗ h ( t ) y(t) = x(t) * h(t) y ( t ) = x ( t ) ∗ h ( t ) The convolution operation is defined as:
y ( t ) = x ( t ) ∗ h ( t ) = ∫ − ∞ + ∞ x ( τ ) h ( t − τ ) d τ y(t) = x(t) * h(t) = \int_{-\infty}^{+\infty} x(\tau)h(t - \tau) d\tau y ( t ) = x ( t ) ∗ h ( t ) = ∫ − ∞ + ∞ x ( τ ) h ( t − τ ) d τ In the discrete case:
c [ k ] = f [ k ] ∗ g [ k ] = ∑ m = − ∞ + ∞ f [ m ] g [ k − m ] c[k] = f[k] * g[k] = \sum_{m = -\infty}^{+\infty} f[m]g[k - m] c [ k ] = f [ k ] ∗ g [ k ] = m = − ∞ ∑ + ∞ f [ m ] g [ k − m ] This operation is also:
Commutative:
f 1 ( t ) ∗ f 2 ( t ) = f 2 ( t ) ∗ f 1 ( t ) f_1(t) * f_2(t) = f_2(t) * f_1(t) f 1 ( t ) ∗ f 2 ( t ) = f 2 ( t ) ∗ f 1 ( t ) Distributive:
f 1 ( t ) ∗ ( f 2 ( t ) + f 3 ( t ) ) = f 1 ( t ) ∗ f 2 ( t ) + f 1 ( t ) ∗ f 3 ( t ) f_1(t) * (f_2(t) + f_3(t)) = f_1(t)* f_2(t) + f_1(t) * f_3(t) f 1 ( t ) ∗ ( f 2 ( t ) + f 3 ( t )) = f 1 ( t ) ∗ f 2 ( t ) + f 1 ( t ) ∗ f 3 ( t ) Associative:
f 1 ( t ) ∗ ( f 2 ( t ) ∗ f 3 ( t ) ) = ( f 1 ( t ) ∗ f 2 ( t ) ) ∗ f 3 ( t ) = f_1(t) * (f_2(t) * f_3(t)) = (f_1(t) * f_2(t)) * f_3(t) = f 1 ( t ) ∗ ( f 2 ( t ) ∗ f 3 ( t )) = ( f 1 ( t ) ∗ f 2 ( t )) ∗ f 3 ( t ) = Using these properties we can say that:
y ( t ) = f ( t ) ∗ h ( t ) ∣ h ( t ) = h 1 ( t ) ∗ h 2 ( t ) y(t) = f(t) * h(t) \ | \ h(t) = h_1(t) * h_2(t) y ( t ) = f ( t ) ∗ h ( t ) ∣ h ( t ) = h 1 ( t ) ∗ h 2 ( t ) In the parallel case:
f ( t ) ∗ h ( t ) ∣ h ( t ) = h 1 ( t ) + h 2 ( t ) f(t) * h(t) \ | \ h(t) = h_1(t) + h_2(t) f ( t ) ∗ h ( t ) ∣ h ( t ) = h 1 ( t ) + h 2 ( t ) Shift:
f 1 ( t ) ∗ f 2 ( t ) = c ( t ) f 1 ( t ) ∗ f 2 ( t − T ) = c ( t − T ) f 1 ( t − T ) ∗ f 2 ( t ) = c ( t − T ) f 1 ( t − T 1 ) ∗ f 2 ( t − T 2 ) = c ( t − T 1 − T 2 ) f_1(t) * f_2(t) = c(t) \newline
f_1(t) * f_2(t - T) = c(t - T) \newline
f_1(t - T) * f_2(t) = c(t - T) \newline
f_1(t - T_1) * f_2(t - T_2) = c(t - T_1 - T_2) \newline f 1 ( t ) ∗ f 2 ( t ) = c ( t ) f 1 ( t ) ∗ f 2 ( t − T ) = c ( t − T ) f 1 ( t − T ) ∗ f 2 ( t ) = c ( t − T ) f 1 ( t − T 1 ) ∗ f 2 ( t − T 2 ) = c ( t − T 1 − T 2 ) Convolution with an impulse:
f ( t ) ∗ δ ( t ) = f ( t ) f(t) * \delta(t) = f(t) f ( t ) ∗ δ ( t ) = f ( t ) We can combine these two:
Convolution with an impulse:
f ( t ) ∗ δ ( t − T ) = f ( t − T ) f(t) * \delta(t - T) = f(t - T) f ( t ) ∗ δ ( t − T ) = f ( t − T ) Differential Equations in the time-domain
As we defined, the output of these systems are the result of two independent cases.
So for the zero-input response:
Q ( D ) y 0 ( t ) = 0 Q(D)y_0(t) = 0 Q ( D ) y 0 ( t ) = 0 The solution for y 0 ( t ) y_0(t) y 0 ( t )
y 0 ( t ) = c 1 e λ 1 t + c 2 e λ 2 t + … + c n e λ n t y_0(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} + \ldots + c_n e^{\lambda_n t} y 0 ( t ) = c 1 e λ 1 t + c 2 e λ 2 t + … + c n e λ n t If repeated roots:
y 0 ( t ) = ( c 1 + c 2 t + … + c r t r − 1 ) e λ 1 t + c r + 1 e λ r + 1 t + … + c n e λ n t y_0(t) = (c_1 + c_2t + \ldots + c_r t^{r - 1})e^{\lambda_1 t} + c_{r +1}e^{\lambda_{r + 1} t} + \ldots + c_n e^{\lambda_n t} y 0 ( t ) = ( c 1 + c 2 t + … + c r t r − 1 ) e λ 1 t + c r + 1 e λ r + 1 t + … + c n e λ n t Also in the case we have complex roots, α ± j β \alpha \pm j \beta α ± j β
y 0 ( t ) = c 2 e j θ e ( α + j β ) t + c 2 e − j θ e ( α − j β ) t = c 2 e α t [ e j ( β t + θ ) + e − j ( β t + θ ) ] y_0(t) = \dfrac{c}{2} e^{j \theta} e^{(\alpha + j \beta)t} + \dfrac{c}{2} e^{-j \theta} e^{(\alpha - j \beta)t} = \dfrac{c}{2} e^{\alpha t} \left[ e^{j(\beta t + \theta)} + e^{-j(\beta t + \theta)}\right] y 0 ( t ) = 2 c e j θ e ( α + j β ) t + 2 c e − j θ e ( α − j β ) t = 2 c e α t [ e j ( β t + θ ) + e − j ( β t + θ ) ] Using Euler’s formula:
y 0 ( t ) = c e α t c o s ( β t + θ ) y_0(t) = c e^{\alpha t} cos(\beta t + \theta) y 0 ( t ) = c e α t cos ( β t + θ ) Example 1 ( D 2 + 3 D + 2 ) y 0 ( t ) = 0 (D^2 + 3D + 2)y_0(t) = 0 ( D 2 + 3 D + 2 ) y 0 ( t ) = 0 λ 2 + 3 λ + 2 = ( λ + 1 ) ( λ + 2 ) = 0 \lambda^2 + 3\lambda + 2 = (\lambda + 1)(\lambda + 2) = 0 λ 2 + 3 λ + 2 = ( λ + 1 ) ( λ + 2 ) = 0 λ 1 = − 1 λ 2 = − 2 \lambda_1 = -1 \newline
\lambda_2 = -2 λ 1 = − 1 λ 2 = − 2 y 0 ( t ) = c 1 e − t + c 2 e − 2 t y_0(t) = c_1 e^-t + c_2 e^{-2t} y 0 ( t ) = c 1 e − t + c 2 e − 2 t Trigonometric Fourier series
ω 0 \omega_0 ω 0
n ω 0 n \omega_0 n ω 0 n n n
We can call the fundamental period for, T 0 T_0 T 0
T 0 = 2 π ω 0 T_0 = \dfrac{2\pi}{\omega_0} T 0 = ω 0 2 π So:
f ( t ) = a 0 + a 1 c o s ω 0 t + a 2 c o s 2 ω 0 t + b 1 s i n ω 0 t + b 2 s i n 2 ω 0 t + … f(t) = a_0 + a_1 cos \omega_0 t + a_2 cos 2\omega_0 t + b_1 sin \omega_0 t + b_2 sin 2\omega_0 t + \ldots f ( t ) = a 0 + a 1 cos ω 0 t + a 2 cos 2 ω 0 t + b 1 s in ω 0 t + b 2 s in 2 ω 0 t + … We can write this more compactly.
a n c o s n ω 0 t + b n s i n n ω 0 t = C n c o s ( n ω 0 t + θ n ) a_n cos n\omega_0 t + b_n sin n\omega_0 t = C_n cos(n\omega_0 t + \theta_n) a n cos n ω 0 t + b n s inn ω 0 t = C n cos ( n ω 0 t + θ n ) Where:
C n = a n 2 + b n 2 C_n = \sqrt{a_n^2 + b_n^2} C n = a n 2 + b n 2 θ n = t a n − 1 ( − b n a n ) \theta_n = tan^{-1} \left(\dfrac{-b_n}{a_n}\right) θ n = t a n − 1 ( a n − b n ) So:
f ( t ) = C 0 + ∑ n = 1 ∞ C n c o s ( n ω 0 t + θ n ) f(t) = C_0 + \sum_{n = 1}^{\infty} C_n cos(n\omega_0 t + \theta_n) f ( t ) = C 0 + n = 1 ∑ ∞ C n cos ( n ω 0 t + θ n ) Exponential Fourier series
f ( t ) = ∑ − ∞ ∞ D n e j n ω 0 t f(t) = \sum_{-\infty}^{\infty} D_n e^{jn\omega_0 t} f ( t ) = − ∞ ∑ ∞ D n e j n ω 0 t C n c o s ( n ω 0 t + θ n ) = C n 2 ( e j ( n ω 0 t + θ n ) + e − j ( n ω 0 t + θ n ) ) C_n cos(n\omega_0 t + \theta_n) = \dfrac{C_n}{2} \left(e^{j(n\omega_0 t + \theta_n)} + e^{-j(n\omega_0 t + \theta_n)}\right) C n cos ( n ω 0 t + θ n ) = 2 C n ( e j ( n ω 0 t + θ n ) + e − j ( n ω 0 t + θ n ) ) ( C n 2 e j θ n ) e j n ω 0 t + ( C n 2 e − j θ n ) e − j n ω 0 t \left(\dfrac{C_n}{2} e^{j\theta_n}\right) e^{jn\omega_0 t} + \left(\dfrac{C_n}{2} e^{-j\theta_n}\right) e^{-jn\omega_0 t} ( 2 C n e j θ n ) e j n ω 0 t + ( 2 C n e − j θ n ) e − j n ω 0 t D n = 1 2 C n e j θ n D_n = \dfrac{1}{2} C_n e^{j\theta_n} D n = 2 1 C n e j θ n D − n = 1 2 C n e − j θ n D_{-n} = \dfrac{1}{2} C_n e^{-j\theta_n} D − n = 2 1 C n e − j θ n f ( t ) = D 0 + ∑ n = 1 ∞ D n e j n ω 0 t + D − n e − j n ω 0 t f(t) = D_0 + \sum_{n = 1}^{\infty} D_n e^{jn\omega_0 t} + D_{-n} e^{-jn\omega_0 t} f ( t ) = D 0 + n = 1 ∑ ∞ D n e j n ω 0 t + D − n e − j n ω 0 t f ( t ) = ∑ − ∞ ∞ D n e j n ω 0 t f(t) = \sum_{-\infty}^{\infty} D_n e^{jn\omega_0 t} f ( t ) = − ∞ ∑ ∞ D n e j n ω 0 t Recall (Fourier transform) F ( ω ) = ∫ − ∞ ∞ f ( t ) e − j ω t d t F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t}\ dt F ( ω ) = ∫ − ∞ ∞ f ( t ) e − j ω t d t f ( t ) = 1 2 π ∫ − ∞ ∞ F ( ω ) e j ω t d ω f(t) = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{j\omega t}\ d\omega f ( t ) = 2 π 1 ∫ − ∞ ∞ F ( ω ) e j ω t d ω f ( t ) ⟺ F ( ω ) f(t) \iff F(\omega) f ( t ) ⟺ F ( ω ) Example 2 f ( t ) = r e c t ( t τ ) f(t) = rect\left(\dfrac{t}{\tau}\right) f ( t ) = r ec t ( τ t ) F ( ω ) = ∫ − ∞ ∞ r e c t ( t τ ) e − j ω t d t F(\omega) = \int_{-\infty}^{\infty} rect\left(\dfrac{t}{\tau}\right) e^{-j\omega t}\ dt F ( ω ) = ∫ − ∞ ∞ r ec t ( τ t ) e − j ω t d t F ( ω ) = ∫ − τ 2 τ 2 e − j ω t d t F(\omega) = \int_{-\frac{\tau}{2}}^{\frac{\tau}{2}} e^{-j\omega t}\ dt F ( ω ) = ∫ − 2 τ 2 τ e − j ω t d t F ( ω ) = − 1 j ω e − j ω t ∣ τ 2 − τ 2 F(\omega) = -\dfrac{1}{j\omega} e^{-j\omega t} \bigg\rvert_{\frac{\tau}{2}}^{-\frac{\tau}{2}} F ( ω ) = − j ω 1 e − j ω t 2 τ − 2 τ F ( ω ) = − 1 j ω ( e − j ω τ 2 − e j ω t τ 2 ) F(\omega) = -\dfrac{1}{j\omega} \left(e^{-j\omega \dfrac{\tau}{2}} - e^{j\omega t \dfrac{\tau}{2}}\right) F ( ω ) = − j ω 1 ( e − j ω 2 τ − e j ω t 2 τ ) F ( ω ) = 2 s i n ( ω τ 2 ) ω F(\omega) = \dfrac{2 sin\left(\dfrac{\omega \tau}{2}\right)}{\omega} F ( ω ) = ω 2 s in ( 2 ω τ ) F ( ω ) = τ s i n ( ω τ 2 ) ω τ 2 F(\omega) = \tau \dfrac{sin\left(\dfrac{\omega \tau}{2}\right)}{\dfrac{\omega \tau}{2}} F ( ω ) = τ 2 ω τ s in ( 2 ω τ ) F ( ω ) = τ s i n c ( ω τ 2 ) F(\omega) = \tau sinc\left(\dfrac{\omega \tau}{2}\right) F ( ω ) = τ s in c ( 2 ω τ ) r e c t ( t τ ) ⟺ τ s i n c ( ω τ 2 ) rect\left(\dfrac{t}{\tau}\right) \iff \tau sinc\left(\dfrac{\omega \tau}{2}\right) r ec t ( τ t ) ⟺ τ s in c ( 2 ω τ ) Transfer function
Say we have a system which can be modeled as:
y ( t ) = f ( t ) ∗ h ( t ) y(t) = f(t) * h(t) y ( t ) = f ( t ) ∗ h ( t ) The same system, but in the frequency domain is therefore:
Y ( ω ) = F ( ω ) H ( ω ) Y(\omega) = F(\omega)\ H(\omega) Y ( ω ) = F ( ω ) H ( ω ) Just as we call, h ( t ) h(t) h ( t ) H ( ω ) H(\omega) H ( ω )
Since, in the most general case, we view our frequency functions as complex functions. We can say that
An input signals spectral component of frequency, ω \omega ω ∣ H ( ω ) ∣ |H(\omega)| ∣ H ( ω ) ∣ ∠ H ( ω ) \angle H(\omega) ∠ H ( ω )
However, since we modify the phase, this means that our output signal may have a different waveform than our input.
Recall (Nyquist-Shannon sampling theorem) A real signal whose bandwidth is limited to B B B F s > 2 B \mathcal{F}_s > 2B F s > 2 B
Recall (Laplace transform) The Laplace transformation is a unilateral transform , which just means it is one-sided.
Given a function that is defined for t ≥ 0 t \geq 0 t ≥ 0 [ 0 , ∞ ) [0, \infty) [ 0 , ∞ )
F ( s ) = ∫ 0 ∞ f ( t ) e − s t d t F ( s ) = L [ f ( t ) ] F(s) = \int_0^{\infty} f(t)e^{-st}\ dt \newline
F(s) = \mathcal{L}[f(t)] F ( s ) = ∫ 0 ∞ f ( t ) e − s t d t F ( s ) = L [ f ( t )] 
