EE 201 EECTRIC CIUITS ECTURE 25 The material overed in this leture will be as follows: Natural, Fored and Complete Responses of First Order Ciruits Step Response of First Order Ciruits Step Response of the Ciruit Step Response of the R Ciruit At the end of this leture you should be able to: Distinguish between the natural, fored, and omplete responses in first order iruits Understand the meaning of the step response Derive differential equations for the step response Solve some first order iruit examples with a step response Apply the initial onditions in iruits with a omplete response Better appreiate the importane of the time onstant of a first order iruit Deide approximately how long a apaitor needs to be fully harged Deide if a first order iruit is D.C. or not D.C. Distinguish between ordinary swithes and make before break swithes. Natural, Fored and Complete Responses of First Order Ciruits: We already know that first order iruits lead to the differential equation (DE): dx() t + ax() t = f () t dt General Solution x() t = e e f() t dt+ Ae at at at For iruits with a natural response the DE is homogeneous f() t = 0
at at at x() t e e = 0dt+ Ae x( t) = Ae at x() t = Ae at [natural response of a first order iruit] () at at x t = e e f() t dt [fored response of a first order iruit] x() t = e e f() t dt+ Ae at at at [omplete response of a first order iruit] Complete Response = Fored Response + Natural Response If the iruit ontains independent soures in general the iruit has a omplete response The following statements an easily be proven for first order iruits: 1) If all the independent soures are D.C. f( t) = onstant = b 2) If all the independent soures are D.C. b x() t = + Ae a omplete response at In this leture and the next leture, we will onsider first order iruits with D.C. independent soures. Step-Response of First Order Ciruits: The step response of a given iruit an be loosly defined as the iruit reation to a sudden appliation of an independent D.C. voltage or urrent soure. Step Response of the Ciruit: Example 1: In the given iruit, the apaitor has zero initial voltage [ v (0) = 0 ]. The D.C. voltage soure V is onneted to the iruit by suddenly losing the swith s at t = 0. a) Find v ( t ) for t 0. b) Plot v ( t ) for t 0. ) Explain the iruit operation. d) How long does it take for the apaitor voltage to reah 99% of its maximum possible value?
Figure 1 Solution: a) The iruit shown is valid for t 0, after losing the swith s et us first derive a DE for v ( ) t : KV V + Ri + v = 0 R ir = i V + RiC + v = 0 V + Ri + v = 0 C dv i = C dt dv V + + v 0 = dt Normalize dv 1 V + v = dt dv 1 V + v = dt V solution 1 t v () t Ae = 1 + [using 1 a = & V b = ] v () t = V + Ae t/ Evaluate A v = V + Ae = V + A= A= V 0/ (0) 0 v t V Ve V e t/ t/ () = = (1 ) for t 0 Figure 2
t/ b) A plot of v () t = V(1 e ) for t 0 is shown. Figure 3 ) Initially v (0) = 0 [unharged apaitor, q(0) = Cv (0) = 0 ] As t inreases v () t also inreases As t v () t V [fully harged apaitor, q( ) = Cv ( ) = CV ] Thus, in this iruit an initially unharged apaitor is being harged. It beomes fully harged when the apaitor voltage reahes the onstant value v ( t) = V. d) The maximum value of v ( t ) is V. Assume it takes t seonds for v ( t) to reah the value 0.99 V (99% of the maximum value) 0 v t = V e 0 ( ) (1 t / 0 v t = V e ) = 0.99V (1 e t / ) = 0.99 t/ () (1 ) 0 t0 e / t0 = 0.01 e / 1.99 = 0 t ln(0.01) = t0 = ln(0.01) t 0 = 4.61 = 4.61τ 5τ Thus, in pratie, the iruit does not need infinite time for value V. v ( ) t to reah the maximum possible Important Conlusions: 1) After about 5τ the apaitor beomes almost fully harged [ v V & q CV ]. 2) After about 5τ the iruit beomes almost a D.C. iruit [sine all urrents and voltages in the iruit beome independent of time, whih an easily be verified].
3) After a long time (t >> τ ), the iruit reahes steady state and all the voltages and urrents in the iruit beome D.C. (i.e. the iruit itself beomes a D.C. iruit). 4) After a long time, the apaitor an be replaed by an open iruit (why?). Figure 4 Example 2: The swith s remains in position (1) for a long time. It is suddenly moved to position (2) at t = 0. Find v ( ) t for: a) t < 0 b) t 0 Solution: a) The iruit model for t < 0 is shown. Figure 5 The independent soure is D.C., does it mean we have a D.C. iruit? This is a ruial question, beause even if all independent soures are D.C. this does not automatially mean that the iruit is D.C. If a D.C. soure is onneted to a first order iruit, the iruit eventually beomes D.C. However, we need to wait a long time (t >> τ ) before the iruit beomes D.C. This iruit must have been onneted at t =, beause the iruit remains unhaged for t < 0 the D.C. soure has been onneted for a long time the iruit is atually D.C.
Figure 6 iruit is D.C. iruit is D.C. replae the apaitor by an open iruit. we do not need a differential equation to desribe its behavior. i R = i = 0 (beause of the open iruit) KV 10 + 5i + v = 0 10 + 5 0 + = 0 v = 10V R v v () t = 10 for t < 0. Figure 7 b) the iruit is shown for t 0 For this iruit, the apaitor has an initial voltage the apaitor is ontinuous). Clearly this is not a D.C. iruit. v (0) = v (0 ) = 10V (beause the voltage aross The apaitor starts to disharge and the voltage aross it eventually reahes zero. We already know the general solution to this iruit when t 0 = 0 : t/ = for t 0 v () t v (0) e v (0) = 10 & = 5 3= 15 [s] /15 v () t = 10e t for t 0
Figure 8 ) the graph of v ( t ) for all t is shown in the figure. Figure 9 Step Response of the R Ciruit: Example 3: The swith s is losed suddenly at t = 0. Find an expression for i ( t) for: a) t < 0 b) t 0 ) Plot i ( t ) and omment on the results. d) Can be replaed by a short iruit a short time after losing the swith? e) Can be replaed by a short iruit a long time after losing the swith?
Figure 10 Solution: a) the iruit for t < 0 is shown. Due to the open iruit i () t = i () t = 0 R i () 0 t = for t < 0 Figure 11 b) the iruit for t 0 is shown. Beause the D.C. soure has just been onneted to the iruit iruit is not D.C. We need to develop a differential equation to desribe the variation for i ( t) KV 4 + 3i + v = 0 R i di & = i R v = dt 4 3 2 di + i 0 + = dt di 3 i 4 dt + 2 = ( 3 a = & b = 4 ) 2 3 4 t 8 2 i () t = + Ae = + Ae 3/2 3 1.5t () i t is ontinuous i(0) = i(0 ) = 0
Using 8 i () t Ae 3 1.5t = + 8 8 i = + Ae = + A= 3 3 1.5 0 (0) 0 8 A = 3 8 8 1.5t 8 1.5t i () t = e = (1 e ) for t 0 3 3 3 Figure 12 ) The graph shows that the urrent through the indutor, whih is initially zero, inreases with time until it reahes a D.C. steady-state value after a suffiiently long time. d) A short time after losing the swith, i ( t ) ontinues to hange with t. annot be replaed by a short iruit. [the iruit is not D.C.] e) A long time after losing the swith, i ( t ) beomes D.C. an be replaed by a short iruit. [in fat the iruit itself beomes D.C.]. Example 4: Figure 13 The swith s is suddenly moved from position (1) to position (2) at t = 0. Find i ( ) t for: a) t < 0
b) t 0 ) Plot i ( t) Figure 14 Solution: a) the figure shows the iruit for t < 0 This iruit must have been onneted at t =. the D.C. urrent soure has been onneted for a long time the iruit is atually D.C. Figure 15 we an replae by a short iruit. CDR 10 i = 8= 8A [all the 8 A urrent passes through the short iruit] 10 + 0 i () 8 t = for t < 0
Figure 16 b) the swith s is a speial type of swith alled make before break swith. It is used to ensure that i ( ) t does not go to zero during the infinitely small swithing time. + Thus: i (0 ) = i (0 ) = i (0) = 8A [If we use a swith similar to the ones shown in the previous iruits + i (0 ) = i (0) = 0, why?] Figure 14 the iruit is shown for t 0 Although the iruit ontains only D.C. soures Ciruit is not D.C. [It will eventually beome D.C. after a long time] et us first derive a D.E. for i ( ) t : KV 24 + 2i + v = 0 R di i= ir & v= 0.2 dt di 24 + 2i + 0.2 = 0 dt di 0.2 2i 24 dt + = Normalize di 10i 120 dt + = b i () t = + Ae a at a= 10 & b= 120 120 i () t Ae 10 10t = + i t () = 12+ Ae 10t Use the initial ondition to evaluate A i Ae A 10 0 (0) = 12 + = 12 + = 8
A = 8 12= 4 10t i () t = 12 4e for t 0 Figure 17 ) the figure learly shows that for t 0, i ( t) varies with time, but after a long time it beomes onstant. Figure 18