0.10 Lecture 11:sinusoidal steady-state (sss) or frequency response

Lecture #11:

SINUSOIDAL STEADY-STATE (SSS) OR FREQUENCY RESPONSE

Motivation:

• Many systems operate in the SSS, e.g., electrical power distribution, broadcast, touch-tone telephone.

• SSS response essentially characterizes the system function.

• SSS response is commonly used to measure the system function.

• A common mode of thinking for signal processing tasks is “filtering.” All physical systems act as filters of their input signals. Filtering is an important signal processing method.

Outline:

• Causal, stable systems
• Sinusoidal steady-state response—the frequency response
• Relation of frequency response to system function
• Bode diagrams

• Signal processing with filters
• Lowpass and highpass filters
• Resonance and bandpass filters
• Notch filters
• Conclusions

A system that operates in the SSS . . . well almost

The touch-tone phone

Dialing consists of pressing buttons on the keypad which has 3 columns and 4 rows. How is the information about which button is pushed coded?

Demo on coding in the touch-tone phone.

Sum of sinusoids

$s(t)=\text{sin}({\mathrm{2\pi f}}_{1}t)+\text{sin}({\mathrm{2\pi f}}_{2}t)$

times a pulse window p(t)

$x(t)=p(t)s(t)$

More on the effect of p(t) later!

I. CAUSAL, STABLE SYSTEMS

The system function of an LTI system, H(s), can be used to categorize systems.

• Causal system $\Rightarrow$ ROC is to the right of the rightmost pole of H(s).

• Causal, stable system $\Rightarrow$ ROC is to the right of the rightmost pole of H(s) and all the poles are in the left-half of the s plane.

More on the definition of stable systems at a later time.

A causal, stable system has the pole-zero diagram shown below.

For s in the shaded region, the response to $x(t)={\text{Xe}}^{\text{st}}$ is the steady-state response $y(t)={\text{Ye}}^{\text{st}}=\text{XH}(s)e^{\text{st}}$

II. THE SYSTEM FUNCTION H(S)

1/ Real and complex poles

H(s) is a complex function of a complex variable s. The plots show |H(s)|.

2/ Effect of a zero

3/ Interpretation of H(s) by pole and zero vectors

H(s) that is a rational function has the form

$H(s)=K\frac{(s-z_1)(s-z_2)\text{.}\text{.}\text{.}(s-z_M)}{(s-p_1)(s-p_2)\text{.}\text{.}\text{.}(s-p_N)}$

H(s) consists of products and quotients of the form $(s-s_k)$ . Each of these terms are vectors in the complex s plane:

$(s-z_1)(s-z_2)\text{.}\text{.}\text{.}(s-z_M)$ are called zero vectors,

$(s-p_1)(s-p_2)\text{.}\text{.}\text{.}(s-p_N)$ are called pole vectors.

=

The vector $(s-s_k)$ points from $s_k$ to s. It can be expressed in polar form as

$(s-s_k)=\mid s-s_k\mid e^{j\text{arg}(s-s_k)}$

We now take

$H(s)=K\frac{(s-z_1)(s-z_2)\text{.}\text{.}\text{.}(s-z_M)}{(s-p_1)(s-p_2)\text{.}\text{.}\text{.}(s-p_N)}$

and express the vectors in polar form

$$ $H(s)=K\frac{\mid s-z_1\mid e^{j\text{arg}(s-z_1)}\mid s-z_2\mid e^{j\text{arg}(s-z_2)}\text{.}\text{.}\text{.}\mid s-z_M\mid e^{j\text{arg}(s-z_M)}}{\mid s-p_1\mid e^{j\text{arg}(s-p_1)}\mid s-p_2\mid e^{j\text{arg}(s-p_2)}\text{.}\text{.}\text{.}\mid s-p_N\mid e^{j\text{arg}(s-p_N)}}$

so that

$\mid H(s)\mid =\mid K\mid \frac{\mid s-z_1\mid \mid s-z_2\mid \text{.}\text{.}\text{.}\mid s-z_M\mid }{\mid s-p_1\mid \mid s-p_2\mid \text{.}\text{.}\text{.}\mid s-p_M\mid }$

And

$\text{arg}H(s)=\text{arg}K+\text{arg}(s-z_1)+\text{arg}(s-z_2)+\text{.}\text{.}\text{.}+\text{arg}(s-z_M)-\text{arg}(s-p_1)-\text{arg}(s-p_2)-\text{.}\text{.}\text{.}-\text{arg}(s-p_N)$

And

$H(s)=\mid H(s)\mid e^{j\text{arg}(H(s))}$

III. FREQUENCY RESPONSE

1/ Relation to system function

Note if

$x(t)=\frac{{\text{Xe}}^{\mathrm{j\omega t}}+{\text{Xe}}^{-\mathrm{j\omega t}}}{2}=X\text{cos}(\mathrm{\omega t})$

then

$\begin{array}{}y(t)=\frac{\text{XH}(\mathrm{j\omega })e^{\mathrm{j\omega t}}+\text{XH}(-\mathrm{j\omega })e^{-\mathrm{j\omega t}}}{2}\\ y(t)=\frac{\text{XH}(\mathrm{j\omega })e^{\mathrm{j\omega t}}}{2}+{\left[\frac{\text{XH}(\mathrm{j\omega })e^{\mathrm{j\omega t}}}{2}\right]}^{}\\ y(t)=\mathrm{2R}\left\{\frac{\text{XH}(\mathrm{j\omega })e^{\mathrm{j\omega t}}}{2}\right\}\\ y(t)=\mathrm{2R}\left\{X\mid H(\mathrm{j\omega })\mid e^{j\text{arg}(H(\mathrm{j\omega }))}{e}^{\mathrm{j\omega t}}\right\}\\ y(t)=X\mid H(\mathrm{j\omega })\mid \text{cos}(\mathrm{\omega t}+\text{arg}(H(\mathrm{j\omega })))\end{array}$