Spring 18 page 1 Lecture 15: Sinusoidal Steady State (SSS) Analysis or Phasors without a Phasor more or less 1. Definition. A signal f (t) is periodic with period T > 0 if f (t) = f (t + T ) ; the smallest such T > 0 is the fundamental period. 2. Memories from the corners of your ECE-201. (a) No transients and I mean it; does anybody want a peanut? Are there transients up ahead? If there are, sinusoids be dead. Inconceivable! (Apologies to Princess Bride) Sinusoids in the steady state are what happens when it s late, in time: K cos(ωt +θ) Stable Circuit M(ω )cos( ωt +ϕ(ω )) (b) Total Recall of Phasors from the mindless Dungeons of 201: K θ Stable Circuit M(ω ) ϕ(ω ) 3. Something NEW for YOU in 202. THE NEW TECHNIQUE Guaranteed to make your evening homework easier and you happier for a longer life, thinner waist line, increased energy, more friends, greater success in your endeavors, all because of the Laplace Transform and no tuition increases : Laplace Transform approach to SSS
Spring 18 page 2 K cos(ωt +θ) Stable Circuit with TF H(s) s= jω M(ω )cos( ωt +ϕ(ω )) where (i) M(ω ) = K H( jω ) (ii) ϕ(ω ) = H( jω ) +θ Question: You re kidding, right? I just have to replace a STABLE (not like in Horses) H(s) by H( jω ) to compute the SSS. That is too easy. It cannot possibly be true not true in any other engineering courses!!!!. ANSWER: Well it is! Why? Because you have had to learn a very sophisticated mathematical technique called Laplace transform analysis, no easy task. You have grown, mathematically, and in circuit theory. One hopes that after all that STRUGGLING WORK something would get easier, RIGHT? THE NOTION OF FREQUENCY RESPONSE: H( jω ) as ω ranges over all frequencies is the so-called frequency response of a circuit or system; (a) H( jω ) is the magnitude response (gain frequency response), and (b) H( jω ) is the phase response. Remark: frequency response is fundamental to much of electrical engineering.
Spring 18 page 3 In regards to frequency response: RULES OF THUMB: (i) Two most important frequencies are s = jω = 0 and s = j ; (ii) The next set of important frequencies are s = jω 1 which makes the angle of the numerator 45 o and s = jω 2 which makes the angle of the denominator 45 o. Example 1. Suppose H(s) = V out = 10 s + 4. For SSS, V in s +100 v in (t) = 10cos(ωt)u(t) H(s) v out,ss (t) = M(ω )cos( ωt +ϕ(ω )) Using trick formulas from complex number theory: (a) M(ω ) = 10 jω + 4 jω +100 10 = 100 jω + 4 jω +100 (b) ϕ(ω ) = H( jω ) + ( Input Signal) = H( jω )
Spring 18 page 4 Table of Values: Approximated ω M(ω ) ϕ(ω ) 0 4 0 o 100 0 o 4 4 2 45 o 100 100 45 o 2 1000 100 0 o 10 9 8 7 Magnitude H 6 5 4 3 2 1 0 0 50 100 150 200 250 300 350 400 450 500 Frequency in rad/s
Spring 18 page 5 70 60 50 Phase H in degrees 40 30 20 10 10 0 10 1 10 2 10 3 Frequency in rad/s Example 2. Find v out,ss (t) for the circuit below when v in (t) = K cos(ωt) V, K > 0. All passive RLC circuits are stable when there are no active/dependent sources. Step 1. Construct H(s). By V-division,
Spring 18 page 6 H(s) = V out V in = 1 s 1+ s + 1 s = 1 s 2 + s +1 Step 2. (i) v out,ss (t) = M(ω )cos( ωt +ϕ(ω )) in which case (ii) M(ω ) = K H( jω ) = M(ω ) = K 1 ω 2 + jω K ( 1 ω 2 ) 2 + ω 2 (iii) ϕ(ω ) = H( jω ) + 0 o = tan 1 ω 1 ω 2
Spring 18 page 7 SSS Analysis Worksheet. Problem 1: The input v in (t) = 10cos(100t 45 o ) V excites a circuit having transfer function H(s) = the steady state output. (100 + s)(100 s) s 2 + s + (100) 2. Find the magnitude and phase of Step 1: What is the frequency of analysis? jω = Step 2: H( jω ) = Step 3: Magnitude M(ω ) = Step 4: Phase ϕ(ω ) :
Spring 18 page 8 Problem 2: Consider the circuit Step 1. Magnitude of v in (t) is:. Angle of v in (t) is: rads. Step 2. After a 201-like source transformation, the circuit becomes a parallel RLC. Add needed labels to the transformed circuit below: Step 3. Find the impedance, Z in (s), seen by the I-source of step 2. Z in (s) = 1 Y in (s) =
Spring 18 page 9 Step 4. Find the transfer function. Note: the actual input is V in (s). So H(s) = V C (s) V in (s) = Step 5. The magnitude of H(s) at s = j2 is: The angle of H(s) at s = j2 is: rads deg Step 6. v C,ss (t) = M cos(2t +ϕ) V. The magnitude M =. Then Angle φ = rads; deg
Spring 18 page 10 Problem 3 (Do this problem before bed time to induce sleep): Find v out,ss (t) = M(ω )cos(ωt +ϕ(ω )) V for the circuit below when v in (t) = 10cos(2t)u(t) V? Step 1: Determine ω. Step 2: Find H(s). Step 3: Find H( jω ) at ω =?? and then M(ω ). Step 4: Find H( jω ) at ω =?? and then ϕ(ω ).