A single balanced switching mixer is shown in Figure 1.
Figure 1. Single balanced switching mixer
MOS transistors 
and 
switch the current from the 
-block between the + and – output terminals. The differential LO signals that drive the switches are shown in the Figure 1.
(1) 
By Fourier series,
(2) 
The output voltage 
of single ended switching mixer is given by,
(3) ![\begin{eqnarray*} v_o(t) &=& v_{o2} - v_{o1} = -g_m R_L v_i(t)~[V_{LO^-}(t) - V_{LO^+}(t)]\\ &=& {4\over \pi} g_m R_L v_i(t) ~ \sum\limits_{n=1,3,5,...}^{\infty}{1\over n}\sin(n\omega_{LO}t)\\ &=& {4\over \pi} g_m R_L (V_b + V_{RF}\cos(\omega_{RF}t)) ~ \sum\limits_{n=1,3,5,...}^{\infty}{1\over n}\sin(n\omega_{LO}t)\\ &=& \begin{cases} {4\over \pi}g_m R_L V_{b}\underbrace{ \sum\limits_{n=1,3,5,...}^{\infty} \sin(n\omega_{LO}t)}_{\text{LO feedthrough}}\\ +{2\over \pi}g_m R_L V_{RF}\underbrace{\begin{cases}\left(&[\sin(\omega_{LO}-\omega_{RF})t+\sin(\omega_{LO}+\omega_{RF})t] \\ & + {1\over 3}[\sin(3\omega_{LO}-\omega_{RF})t+\sin(3\omega_{LO}+\omega_{RF})t]+\ldots \right) \end{cases}}_{\text{mixed components}} \end{cases} \end{eqnarray*}](https://analog.intgckts.com/wp-content/ql-cache/quicklatex.com-75ca8a4c3c315c7aa006b1f7b7699229_l3.svg "Rendered by QuickLaTeX.com")
From Eq.(3), we can notice RF feedthrough suppression by differential LO signaling. Figure 2 illustrates the output voltage spectrum of a single balanced switching mixer.
Figure 2. Single balanced switching mixer output voltage spectrum
Apart from sum and difference components due to mixing action, LO feedthrough still exists which is due to DC component in RF signal. This can be eliminated through differential RF signal in double balanced switching mixer.