Can you help me with these equations based on the given circuit? 11. Calculate the low frequency cutoff values due to the 3 external capacitances. Show yourwork and record the values in Table 1.a. Frequency due to the input capacitor. The source resistance in this case is inseries with Rin(tot)fel2T (600+fez =2n(Rs+(R₂||R₂|| (Pac(r'e+RE1))))C₁fc3 =11168000 22000 94.6(1.38+33).b. Frequency due to the bypass capacitor. The source resistance in this case is inparallel.12π(R₂ +R₂) C₂2n(RE₂r'e+RE+R₂R₂ KS)C=1(100-10-6)Bac= 479.36c. Frequency due to the output capacitor.Equation 3= 4.571 Vs10mVpk5kHz0°2.R7m6000RsC1DE100nFR168kQR2Σ22kΩVCC15.0VR33.9KQRcQ12N3904R4330RE1R5>1.5kΩRE2+₁C3DE100uFC2100uFR635.6kΩRLH1₁

Given:(a) The lower cut-off frequency due to the coupling capacitor C1 is given as, fLC1=12πRs+RinC1,Here, Rin is the effective resistance seen from the capacitor C1 or it is known as the input impedance of the BJT and Rs=R7.Rin=R1∥R2∥βr'e,fLC1=12πRs+R1∥R2∥βr'eC1 1 Now, in part a value of β=βac=94.6. In order to calculate the small-signal emitter resistance re, the DC analysis of the BJT is performed as given below, The voltage across the base is given as, VB=VCC×R2R1+R2,VB=15×2268+22,VB=3.67 Volts. The current through the emitter terminal is given as, IE=VEREIE=VB-VBE33Ω+1500Ω In DC analysis both resistors RE1 and RE2 will be considered as capacitor is open-circuitedIE=3.67-0.71533,IE=1.9352 mA. The small-signal emitter resistance is given as, re=VTIE,re=26 mV1.9352 mA,re=13.4353 Ω.Effective small-signal resistance will be,r'e=re+RE1 Since, RE2 will be bypassed by capacitor C2 in the AC analysisr'e=13.4353+33,r'e=46.4353 Ω. Substituting the required values in equation 1, fLC1=12πRs+R1∥R2∥βr'eC1,fLC1=12π600+1168000+122000+194.6×46.4353×100×10-9,fLC1=390.6 Hz.(b) The lower cut-off frequency due to the bypass capacitor C2 is given as, fLC2=12πReffC1 2Reff=RE2∥R'sβ+r'eHere, R's=Rs∥R1∥R2,R's=11600+168000+122000=579.0967 Ω,Reff=RE2∥R'sβ+r'e,Reff=11RE2+1R'sβ+r'e,Reff=111500+1579.096794.6+46.4353,Reff=50.77768 Ω. Substituting all the required values in equation 2, fLC2=12π×50.77768×100×10-6,fLC2=31.34348 Hz.(b) The lower cut-off frequency due to the coupling capacitor C3 is given as, fLC3=12πRC+RLC3,fLC3=12π3.9 kΩ+5.6 kΩ×100×10-6,fLC3=0.16753 Hz. Note: In the given question, the solution of r'e is wrong it should be 46.4353 Ohms as in DC analysis both the resistors at the emitter terminal will be taken into account. This is the reason the value of lower cut-off frequencies calculated in all the steps is different from the given solution.