#20 Stable causal sinusoidal response, RC low-pass magnitude and phase, and amplitude versus power decibels

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. Evaluate the stable causal transfer function on the imaginary axis

   

   For the chosen initially relaxed LTI circuit:

   $\displaystyle Y\left(s\right)=X\left(s\right)H\left(s\right)$

   Keep the same chosen input-output map.

   Now ask how each sinusoidal frequency is changed by a stable causal system.

   When the imaginary axis belongs to the transform region, evaluate there:

   $\displaystyle s=j\omega$

   Narration transcript

   In the previous lesson, the transfer function was our zero-state map: Y of s equals X of s times H of s. Now we use that same map in a more practical way. Instead of asking for every possible time waveform, we ask how the circuit treats each sinusoidal frequency. The move is beautifully small: take H of s and evaluate it on the imaginary axis, by setting s equal to j omega.

2. 2. Read frequency-dependent gain and phase

   

   For a stable causal LTI circuit, its frequency response is the transfer function evaluated on the imaginary axis.

   At each angular frequency it has a complex gain.

   Its magnitude gives amplitude gain; its argument gives the phase shift.

   A phase lead or lag of one sinusoid is not automatically a frequency-independent time delay.

   Narration transcript

   Frequency response is H of j omega. It is a complex number for every angular frequency omega. The magnitude tells us the gain at that frequency, and the angle tells us the phase shift. So the circuit becomes a frequency-by-frequency filter: some frequencies pass almost unchanged, some are attenuated, and each one can be delayed or advanced in phase.

3. 3. Separate steady state from the transient

   

   Use a real stable LTI circuit and wait for the transient to decay.

   For sinusoidal input, the steady-state output is:

   $\displaystyle x\left(t\right)=A \cos \left(\omega t\right), y_{\mathrm{ss}}\left(t\right)=A|H\left(j\omega \right)|\cos \left(\omega t+\varphi \right)$

   The steady-state frequency is unchanged.

   The gain and phase change; φ denotes the argument of the frequency response.

   This rule gives the steady-state part, not the complete switched-on waveform.

   Narration transcript

   Here is the key steady-state rule. If the input is A cosine omega t, then after transients die out, the output is A times the magnitude of H of j omega, cosine of omega t plus the angle of H of j omega. The frequency stays the same. Only the amplitude and phase change. That is why frequency response is so powerful for sinusoidal analysis.

4. 4. Apply the RC low-pass transfer function

   

   For the RC voltage low-pass with positive constant R and C:

   $\displaystyle \tau =R C, H\left(s\right)=\frac{1}{\tau s+1}$

   Its stable causal transform region contains the imaginary axis:

   $\displaystyle s=j\omega$

   The frequency response is:

   $\displaystyle H\left(j\omega \right)=\frac{1}{1+j\omega \tau }$

   Well below the corner, the dimensionless product is small:

   $\displaystyle \omega \tau \ll 1, H\left(j\omega \right)\approx 1$

   Well above the corner, the voltage gain is small; this assumes the same input amplitude.

   Narration transcript

   Use the same RC low-pass transfer function from last time: H of s equals one over tau s plus one, where tau equals R C. Frequency response means substitute s equals j omega. So H of j omega equals one over one plus j omega tau. At low frequency, omega tau is small, and the denominator is close to one. At high frequency, omega tau is large, and the output becomes small.

5. 5. Read the complete magnitude and lag formulas

   

   The entire dimensionless product is squared:

   $\displaystyle |H\left(j\omega \right)|=\frac{1}{\sqrt{1+\left(\omega \tau \right)^{2}}}$

   For positive frequency the RC phase is negative:

   $\displaystyle \varphi \left(\omega \right)=-\arctan \left(\omega \tau \right)$

   At very low positive frequency the gain tends to one and the phase tends to zero.

   At the corner, in degrees:

   $\displaystyle \omega _{c} \tau =1, \varphi \left(\omega _{c}\right)=-45^{\circ}$

   At very high positive frequency the phase approaches minus ninety degrees.

   Narration transcript

   From H of j omega equals one over one plus j omega tau, the magnitude is one over the square root of one plus omega tau squared. The phase is minus arctangent omega tau. At very low frequency, magnitude is almost one and phase is almost zero degrees. At the corner, omega tau equals one, the phase is minus forty-five degrees. At very high frequency, phase approaches minus ninety degrees.

6. 6. Keep amplitude and power ratios distinct

   

   A logarithmic scale makes a wide range of positive magnitude ratios readable.

   For a dimensionless voltage or current amplitude ratio r, the numerical gain in decibels is:

   $\displaystyle G_{A}\left(r\right)=20 \log _{10}\left(r\right)$

   Unity amplitude ratio gives zero decibels:

   $\displaystyle r=1, G_{A}\left(1\right)=0$

   This amplitude ratio gives approximately minus three decibels:

   $\displaystyle r=\frac{1}{\sqrt{2}}, G_{A}\left(r\right)\approx -3.0103$

   One tenth and ten times amplitude give opposite gains in decibels:

   $\displaystyle G_{A}\left(0.1\right)=-20, G_{A}\left(10\right)=20$

   For a positive power ratio use:

   $\displaystyle G_{P}=10 \log _{10}\left(\frac{P_{2}}{P_{1}}\right)$

   Narration transcript

   Engineers usually plot amplitude ratios in decibels. For voltage or current amplitude ratios, d B equals twenty log base ten of the magnitude ratio. A ratio of one is zero d B. A ratio of one over square root of two is about minus three d B. A ratio of point one is minus twenty d B, and a ratio of ten is plus twenty d B. For power ratios, use ten log base ten instead.

7. 7. Distinguish the exact RC curve from its asymptote

   

   Use a logarithmic positive-frequency axis and amplitude gain measured in decibels.

   The angular corner frequency is the reciprocal time constant:

   $\displaystyle \omega _{c}=\frac{1}{\tau }$

   At the corner the exact RC gain in decibels is:

   $\displaystyle G\left(\omega _{c}\right)=-10 \log _{10}\left(2\right)\approx -3.0103$

   Well below the corner, the exact curve approaches zero decibels.

   Far above the corner, the slope approaches minus twenty decibels per decade; it is not exactly that slope just above the corner.

   Narration transcript

   The Bode magnitude plot puts frequency on a log axis and gain in decibels. For the RC low-pass, the corner frequency is omega c equals one over tau. At that corner, the gain is minus three point zero one d B. Before the corner, the curve is close to zero d B. After the corner, it falls at about minus twenty d B per decade.

8. 8. Keep units, impedance assumptions and phase signs

   

   Four checks before reading a frequency-response plot.

   Angular frequency is in radians per second and ordinary frequency in hertz:

   $\displaystyle \omega =2\pi f$

   Use twenty log for amplitude and ten log for power; equal-resistance conditions are needed to identify the amplitude result with the power gain.

   For positive frequency this RC voltage low-pass has a negative phase: it lags.

   This sinusoidal rule assumes decaying transients; an unstable or imaginary-axis-pole system need not settle.

   Narration transcript

   Four common traps. First, omega is angular frequency in radians per second; ordinary frequency f is in hertz, and omega equals two pi f. Second, use twenty log for voltage or current amplitude ratios, but ten log for power ratios. Third, phase signs matter: this RC low-pass lags, so its phase is negative. Fourth, frequency response describes sinusoidal steady state after transients have died out.

9. 9. Use the RC frequency-response landmarks

   

   Summary: gain and phase are frequency-dependent properties.

   Start with the specified zero-state input-output transfer function.

   For a stable causal system, evaluate it on the imaginary axis:

   $\displaystyle s=j\omega$

   Keep amplitude gain, phase and logarithmic gain distinct; phase need not be a constant time delay.

   RC corner landmarks:

   $\displaystyle \omega _{c}=\frac{1}{\tau }, |H\left(j\omega _{c}\right)|=\frac{1}{\sqrt{2}}, \varphi \left(\omega _{c}\right)=-45^{\circ}$

   Next: compare low-pass and high-pass magnitude, phase and Bode behavior.

   Narration transcript

   Summary. Start from the transfer function H of s. Set s equal to j omega to get frequency response. The magnitude gives frequency-dependent gain, the angle gives phase shift, and decibels make wide ratio ranges easy to read. For the RC low-pass, the corner is one over tau, the gain is minus three d B there, and beyond it the magnitude falls at about minus twenty d B per decade. Next we can compare low-pass and high-pass filters using the same language.

Source video: Circuit Theory-2 #20 | Frequency Response and Decibels (4:50)