Question

PLZ answer both parts INCLUDING LTSPICE PORTION thanks

IdealXfrmZin02

a) manually find, draw, and label the impedance, Zin, seen at terminals a,b for the following frequencies ............... ............ ..............

IdealXfrmZin02 mo L N:N2 w C Zin Given: The transformer is ideal, well duh. Do: a) Manually find, draw, and label the impedance, Zn, seen at terminals a, b for the following frequencies: i. f = 1 ii. f = f The "draw" part should appear as one or the other of the following: ? j? OR b) Use SPICE to simulate the circuit above with an independent current source as the input to the terminals a, b. Set the current source AC magnitude equal to one and the AC phase equal to zero. Run with an AC Sweep (5000 points per decade) between f = f, and f = f2. Plot the voltage between terminals a, b. Change the vertical magnitude axis from "decibel" to “linear”. You will see the left vertical axis units change from "dB" to "V". Use the cursor feature to find the resonant frequency, f,, which lies between the 2 frequencies cited above: fi<f,<f, where the input impedance is purely real. The input impedance, Z., , will be the voltage between terminals a, b. ... Duh : Annotate the location and value off, on the printout of your plot. Use manually drawn arrows, text, etc. to indicate f. Annotate the location and value of Z) = R + j0=R, directly on the plot printout. It will be found on the vertical axis. You can get somewhat accurate values off, and Z from the pop-up box for the attached cursor. You'll know that the cursor crosshairs are close to ſ, when the box with the phase value shows approximately zero. The cursor pop-up doesn't print when you print the plot.

Answer 1

Part a)

229.7204HH I N: Nz + 37:93 c = 6.092942,UF R Vio t at f = f = 2kHz A Zin 1. L Vi side- Referring capacitor to c = c * (Ni 2 MF c'= 6.09 2942 x 93 0.9644 MF R www equivalent Network Zeq = (R!! + ما به ز + ( عنه a -jp/wc' +jWL 11 --50 tjwL Zeq = Rj/we -jr wreel At RkHz 21x2000 = 12566.3706 rad/s. +></2sz,3704 x 229.7204x10-4 zer 88675 -34 12566.3706x4x0.9644X10-6 J +j 208 0.04847 ju (0.04847 +j) Zer Zeq Zeq +j2 2-886 75 1.0023 =j 0.1934 +3.9908 +j2.80675 Zeze tj.Z.69335 + 3.9908) 32.6333525 3.99oor At f = 72 = 6kHz 2T X 6000 = لا 3 7699. 1118 Mad / ser, in eq equis Sulting Zeq of j 37699.1118x229,7204710-6 Zeq -jy 37639.1118 * Y x 0.9644x10-6 j -jy tj 8.66 0.14543-) j10.14543 tj) +39.64 Zeq - 1.021) Zeq = / 8.51758 +0.979322 0.97934 38.5175851 j8.51758. b a

Part b)

Schematic:

As Number of Turns are directly proportional to (Inductance)^2, we set primary inductance to N1^2 and secondary inductance to N2^2 to get appropriate transformer,

L3 0 0 229.7204 k L1 L2 1 R1 C1 L1 L2 4 6.092942 8649 1369 b OLI 11 iac dec 5000 2K 6K AC 1

Frequency response-

-V(b) 8.4V- 90° 7.8V 6KH7,7.2392311V +80° 7.2V- 70° 6.6V- 60° 6.0V- 50° 5.4V- 40° 4.8V- 30° 4.2V- 20° 3.6V- 10° 2.0073892KHz,2.12808V 3.0V- 0° 2.44 -10° 2:509377KHz,1.4571613V 1.8V- -20° 1.2V- 2.0KHz -30° 6.0KHZ 2.4KHz 2.8KHz 3.2KHz 3.6KHz 4.0KHz 4.4KHz 4.8KHz 5.2KHz 5.6KHz

Hence -

Resonance freq. = 2.509 kHz

Re(Z) = 1.45716 Ohm at resonance