Complex Numbers, Signals, and Circuits

Transcription

1 Complex Numbers, Sgnals, and Crcuts 3 August, 009 Complex Numbers: a Revew Suppose we have a complex number z = x jy. To convert to polar form, we need to know the magntude of z and the phase of z. z = zz = x jyx jy = x y { y θ = arctan { x} } y = arccos z { } x = arcsn z Example : Convert j nto polar form. Frst, note that Re{ j} =, Im{ j} =. We use the above formulas to fnd the modulus magntude and phase of j.. Therefore, j = = { } θ = arctan = π 4 j = { exp j π }. 4 Example : Wrte j j n cartesan form. To solve ths problem, we use the fact that j = e j π. Thus, j j = e j π j, = e j π, = e π.

2 Basc Sgnals In class, we learned about the unt step functon and the pulse functon. Here we wll examne two other contnuous sgnals that are related to the functons already presented, as well as dscrete-tme functons.. Contnuous Sgnals Frst we consder the square wave. sqt Δ t Fgure : Ths depcts the square wave functon, sqt. Δ The square wave s a perodc, contnuous functon. It s perod s Δ, whch means that the frequency s. We can re-wrte the square wave usng a sum of delayed pulse sgnals: sqt =... p Δ t Δ p Δ t p Δ t Δ... Another mportant sgnal s the ramp sgnal. The ramp functon s zero for all negatve values of t. For rt t Fgure : Ths depcts the ramp functon, rt. t > 0, the functon smplfes to rt = t. Therefore, we can wrte the ramp functon n terms of the unt step functon: rt = tut. We note that the unt step functon ntegrates to the ramp functon.. Dscrete-Tme Sgnals The basc tool used to sample contnuous-tme sgnals, or create dscrete-tme sgnal representatons, s the unt sample functon a.k.a. the mpulse functon.

3 δn n Fgure 3: Ths s the unt sample functon mpulse, δn. We can use the unt sample functon to decompose other dscrete-tme sgnals. For example, consder the dscrete-tme sgnal πn sn = cos. 8 Ths sgnal s a sampled verson of the contnuous-tme sgnal cost and has samplng frequency f s = 8. The above dscrete-tme sgnal can be wrtten as πt st = cos δk t. 8 k= Any dscrete sgnal can be wrtten as a sum of unt samples delayed and scaled. 3 RLC Crcuts Recall our basc crcut elements: resstor, nductor, and capactor. R v Fgure 4: Resstor We relate current, voltage, and resstance wth the followng equaton: vt = Rt. C v Fgure 5: Capactor We can relate the capactance, C, n terms of the current and voltage as well. t = C dvt dt 3

4 L v Fgure 6: Inductor vt = L dt dt Generally we are gven the values for R, L, and C, and we wsh to solve for vt or t. For now, all voltages and currents wll be constant,.e. unchangng wth tme. Defnton: a open crcut occurs when no current flows but there may be a non-zero voltage. Ths s equvalent to a resstor havng a value R. Defnton: a short crcut occurs when voltage goes to zero for a possbly non-zero current flow. Ths s equvalent to a have resstor havng a value R 0. Example 3. R 3 v s ± v v v 3 R 3 4 For ths example, we wsh to solve for the current passng through the resstor, 3 the current passng through the resstor R 3, and v out, whch for our purposes wll be the voltage V 3. We begn by wrtng the KCL equatons: = 3, 3 = 4. We can see from the crcut dagram that 4 =, so the second KCL equaton becomes 3 =. 3 Now we wrte the KVL equatons. Note that ths crcut has three closed loops, but t turns out that the system of the three loop equatons s lnearly dependent. We only need to wrte equatons for two of the closed loops. Here they are: v v v s = 0, 4 v 3 v =

5 For each resstor we also have a v relaton: v = R, 6 We use 7 and 8 to rewrte 5. v =, 7 v 3 = 3 R R 3 = 0 3 R 3 = Usng n ths last equaton, we have 3 R 3 = 3 3 R 3 = R3 3 = If we add together 4 and 5, and substtute n the v relatons, we get We now rewrte n 9 n terms of 3. R3 3 R 3 R 3 v s = 0 R3 R R 3 3 = v s 3 = v s R 3 R 3 v s = 0 9 R 3 R R 3 We have solved for one our objectve quantes, 3. Note that as ncreases, the current 3 ncreases; ths means that as resstance for ncreases, the current decreases and 3 ncreases. Current prefers to travel a path of lower resstance. Usng our soluton for 3 and 8, we also know v out : v out = v s R 3 R 3 R R 3 R 3 = v s R 3 R R 3 We can solve for n a smlar fashon as we solved for 3. Ultmately, we fnd R 3 = v s. R 3 R R 3 Here, agan, we see that f R 3 s ncreasng, the current through wll ncrease. 5