Question

Question 4. Refer to the circuit of Figure 4. R 802 50 uF с vi(t) v.(t) Figure 4 a) Draw the circuit in the Laplace domain, and then apply basic circuit theory in the Laplace domain to show that the Laplace transfer function H(s) defined for this system is: HS) V.(5) V (5) sta where a= RC [8 Marks] b) Use Laplace methods to determine the output voltage vo(t) when the input voltage is defined as: v (1) 40(1) The values of R and C are shown in the circuit diagram, and you are allowed to use Laplace tables. [12 Marks)

Answer 1

Solution:

Key points:

By the observations we can say the given circuit can work as a LOW PASS FILTER .

Because at lower frequency capacitor offer infinite resistance hence it will be open circuited so the Output voltage will be same as the input ( Vo = Vin ) so it signifies it is passing all the lower frequencies.

For higher values of frequency capacitor will be short circuited hence output voltage will be zero so Vout = 0V . Hence its attenuating the higher frequencies hence its a LOW PASS FILTER.

Transfer Function (H(S)) = Vo(S) / Vi (S).

Vo(S) = H(S) × Vi(S) . By the above analysis we can get Vo(S) .

Further by applying inverse Laplace transform we get Vo(t).

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4 R un + + Vilt) c V.(+) Laplace transformations Vilt) Laptate Transforma Vi (8) LT LI R LT с SC LT Volt Vols) Replacing all the above Obs es vations in the given circuit diagram R=80N ДАЛА ti + Vols) V (2) 8C Laplace transform equivalent cursuit

R 401 + Vio + YOS SC Ву the voltage write Vols as division sule we can Vo(s) = Vi (8) x 'Isc. Rt 11sc Vils) Iso (R$e+ Ulse Vols) Vis RCS + V.18) - Vils RC ( 5+ RC The transfer function is defined as Vots) H18) : y($)

(8) = Vols) Vils Vols) = V1018) RCUIT TRC Vols) viisi - stre HIRO HIS st/Rc mentioned to take a= RC G H18) sta The transfer function is a H(S) sta a = I RC RC = Time constant for RC circuit

(6 output Vollage Volt) In the question 4 (a) we got transfer function H() is the H68) - Vols) Vi (8) Here Input signal Vilt) - 4 uID Viit. 4 ult) Taking Laplace Transform on both side Vils 4 X I I Villa 4 S Note : ult) La place ol- Transform unit step signal Laplace 4 S 4 ult) 1 Transform Inverse Laplace a Tronstorm -at e uit) sta

H(S) = Vo(s) Vits) Vols HIS Vils a H18) - sta a = '/RE Vo(8) = HV) x Vi (8) on substituting H18) and Vils) valuer Vo(8) = * Vi (3) = 4 a X 4 sta S S Here a = 1 RC In given the question R = 800 they (=Sour RC = 80x sox106 RC = 4000xio RC = 4X10 sec RC = ux103 But we have RC

RC = Ux103 RC 1 -3 4810 3 a = TO 4 a = 1000 4 a = 250 11 Vo(s) = x 4 X a=250 sta S Vols) = x 4 C# 850 I 1000 Vols) S(S+250) partial fraction method Vols) B + А S St250

Vols 1000 A = 0+250 Neglects term A = 1000 210 A 4 Vots too0 B -256 s =-250 Neglect S4250 tem B- -4 V.13) 4 4 S St250 for volt) we must take inverse Laplace Алоидрол н ILT 1,057 ☆ VoH) Volti - 250€ = 4 ult) - 4 e ult Volt) - 4- 4 e 250t) ult) - 250t Volti 4 ult) V