A palindrome is a number or a text phrase that reads the same backwards as forwards. For example, each of the following five-digit integers is a palindrome: 12321, 55555, 45554 and 11611. Write a C++ program that reads in a five-digit integer and determines whether it is a palindrome .

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https://www.chegg.com/homework-help/questions-and-answers/q2-repeat-analysis-q1-e-network-fig-addition-source-resistance-rs-signal-source-vs-shown-f-q80745709

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Solution:

a) the resistance r_e is same with and without source.

b) The mid-band gain without source is A_vmid =-72.754 and with source A_vs = -54.49

c) Input impedance is the same with and without source.

d) Low cut-off frequency due to source, coupling, and bypass capacitor.

Without source the Low cut-off frequency due to the source capacitor is dominant.

With source the Low cut-off frequency due to bypass capacitor is dominant.

e) Low-frequency plot is shown in the uploaded image.

The Calculation part is shown in the uploaded images.

Solution Given 1 1 (s = 0.47 4F (€ = 2014F Ce=0.47HF Ri=68kn, R2 = 10kh, Rc = 5.6kh, RE= 1.2ks, Re = 3.3kr B=120, Vcc= 14 V Rs = 0.82 km2, VEE=0.7V Calculation without Rs and signal source Vs a) To determine re for Dc Conditions Using approximate analysis BRE > 10R2 120 x 162X.103 > 10 x 10 x103 144 x103 > 100x10 The result is 10x10² x 14 Ball voltage VB = Rg Vcc Ritra = 179+V Tox 103 +68x10? The Emitter Current VE IE RE 1794-0.7 0.911mA 10 2x103 26x103 3 26 28:54 0.911xio The resistance re *** IE Midband gain Vo Armid = Roller re -5.6x10|| 3-3x10 72.754 28.540 o The input impedance (27) Z1 = Rp a Rillrall Bre Bre 120 X 26.4 = Zi - 68x103 || 10x103 || 3-42uxiol 2.458 km

d) 2 calculation of Lower cut-off frequency Cutoff frequency due to Source Capacitowice (s Ri = RillRailbre 1 fis 21 Rics 204 58kr. 1 & X 2.458X103 x 0.47x166 Cs fus = 137. 76 Hz 1 Vi & Rille, Bre J Fig: Equivalent circuit for is Cut-off friquenly due to coupling capacitare se. (c) Cc fie T 5 T ㅠ }R, vo 2IT(R(TRL) CC Rc YO . dir (s.6x10°+3.3 *10%)x0-47xjós 38.048 HZ Fig: Equivalent circuit for & 2 Cut off frequency due to Bypass Capacitana (Rille2]+ 1 FLE ZIT RECE 1.263 RE CEOZOKF Re = RE tre B (68k 11 lok - 1.2x10 t 28.54 120 Re= 93.3132 RillR2 1110k + 28.54) 2x 103 111 1166** 1 6 11 FLE & IT Re CE 211 x 93.313 x 20 x 10 = 85.274 HZ

e) Av 3 FLE=85.27 Hz Armial do +22=38.042 fus= 137.76 Midband livel. . -3+ 100 -64 Frequency -388 41 2003 -127 اکرا۔ -18 -21 -247 -21 - Fig: Low frequency plot for the given Circuit Calculation of low frequency - 38.04H2 FLE= = 85.27H2 fis = 137.76H2 Low Cut-off frequency is dominant due to fus = 137.76H2. Calculation with Rs and signal source Vs fuc 1 is same @ The resistance re = 28.542 b) Midband gain Vo - 72,754 Armid = vi Rs The input voltage. vi Vis Rivs Ri Ritrs y 2-458x103 Vi Vs - 0.7498 2.458X103 +0.82x107 fig: With Rs fus The Gain with source vs Vo Ars = Vo Vi - 6- 72-754)x0.7498 VS 2-54.49 Gain Reduces Armid Avs

In but impedance (i): No change in input imbedance it is same 2.458 kr d) Lower cut off frequency Source Capacitor is 1 Jis = La Ceit Rs.cs 27(2-456X163 +0+62 813)X0-47x16 tus 103.303 H2 Coupling Capacitor fuc LIT CRc+R() (c 38.048 HZ Lower cut off frequency due to cc. No change in Bypass capacitor 1 FLE = & IT Rece Where Re = 1 Rs ß ( 1449.5 +28 54 Rs = Rsl Rillea - 0.82x10? 11 68x18? ! cxid? - 810.229 #oxic) Re = 1.2k G 28:54 120 - sto 2k11 34.785 749.52 33.gos - 235.401 H2 } FLE } - 6 21 X 33.8 X 20x10 x 2T ReCE e The Low-Cut off frequency fus = 103.303 42, fuc= 35.048H2 FLE=235.401 42 The low cut off frequenly is dominant due to fLE = 235-401 H2