Question 8.7: A FET transcharacteristic can be approximated as a third−ord......
A FET transcharacteristic can be approximated as a third-order polynomial as:
i_D(t)=a_1 e_G+a_2 e_G^2+a_3 e_G^3 .
Derive the corresponding descriptive function model around a carrier frequency f_0 and show that the model correctly predicts the in-band third-order intermodulation products and also the two-tone output current at the fundamental.
Since the starting model is memoriless, the carrier frequency is immaterial. Starting from a single-tone test where e_G=E_G \cos \theta, we have already derived the expression of the fundamental output current as:
i_{D, ib }(t)=I_{D, ib } \cos \theta=\left(a_1+\frac{3}{4} a_3 E_G^2\right) E_G \cos \theta,
where i_{D, \text {ib }} denotes the frequency components of the output current close to the carrier, i.e.:
I_{D, ib }=F_P\left(E_G\right) E_G, \quad F_P=a_1+\frac{3}{4} a_3 E_G^2 .
Notice that, due again to the memoriless nature of the input-output relationship, the phasor is real and there is no phase delay between the input and the output. Moreover, the descriptive function does not include the second-order term of the power series, but only the odd-order terms. Let us now introduce a two-tone input signal, e_G(t)=E_G \cos \theta_1+E_G \cos \theta_2. We can write:
e_G(t)=\operatorname{Re}\left[E_G e ^{ \text{j} \theta_1}+E_G e ^{ \text{j} \theta_2}\right]=\operatorname{Re}\left[E_G\left(1+ e ^{ \text{j} \Delta \theta}\right) e ^{ \text{j} \theta_1}\right] ,
with \Delta \theta=\theta_2-\theta_1, so that:
Taking into account that:
\left|1+ e ^{ \text{j} \Delta \theta}\right|^2=\left(1+ e ^{ \text{j} \Delta \theta}\right)\left(1+ e ^{- \text{j} \Delta \theta}\right)=2+ e ^{ \text{j} \Delta \theta}+ e ^{- \text{j} \Delta \theta},
we obtain:
Recovering the time-domain output current (remember that \theta=\omega t):
i_{D, ib }(t)=\operatorname{Re}\left[I_{D, ib }(t) e ^{ \text{j} \theta_1}\right] ,
we have:
Thus:
i.e., comparing with (8.12),
i_D(t)=I_D^{(1)} \cos \theta_1+I_D^{(2)} \cos \theta_2+I_D^{(+)} \cos \left(2 \theta_2-\theta_1\right)+I_D^{(-)} \cos \left(2 \theta_1-\theta_2\right) \hspace{30 pt} \text{(8.12)}
we have the same result as in (8.13).
\begin{aligned}I_D^{(1)} & =\left(a_1+\frac{9}{4} a_3 E_G^2\right) E_G \;\;\;\;\;\hspace{30 pt} \text{(8.13a)}\\ I_D^{(2)} & =\left(a_1+\frac{9}{4} a_3 E_G^2\right) E_G \hspace{30 pt}\;\;\;\;\;\text{(8.13b)}\\ I_D^{(+)} & =\frac{3}{4} a_3 E_G^3 \hspace{30 pt}\hspace{30 pt}\;\;\hspace{30 pt} \text{(8.13c)}\\ I_D^{(-)} & =\frac{3}{4} a_3 E_G^3,\hspace{30 pt}\hspace{30 pt}\hspace{30 pt} \text{(8.13d)} \end{aligned}