A single ended switching mixer is shown in Figure 1.
Figure 1. Schematic of a single ended switching mixer
The Linear 
-block convert the RF signal from voltage to current domain.
The switch(S) is turned ON and OFF by the LO signal. LO signal is a square wave with 50% duty-cycle.
(1) 
By Fourier series,
(2) 
When the switch is driven by LO, the RF signal gets multiplied by square wave … changes sign.
Therefore output voltage (
) is given by,
(3) ![\begin{eqnarray*} v_o(t) &=& V_{DD} - g_m R_L v_i(t) V_{LO}(t) \\ &=& V_{DD} - g_m R_L (V_{b} + V_{RF} \cos\omega_{RF}t) \left( {1 \over 2} + {2\over\pi} \sum\limits_{n=1,3,5,...}^{\infty} \sin(n\omega_{LO}t) \right) \\ &=& \begin{cases} \underbrace{V_{DD} - {1\over 2}g_m R_L V_{b}}_{\text{DC component}} \\ - {1\over 2}g_m R_L \underbrace{V_{RF} \cos\omega_{RF}t}_{\text{RF feedthrough}} \\ - {2\over \pi}g_m R_L V_{b}\underbrace{ \sum\limits_{n=1,3,5,...}^{\infty} \sin(n\omega_{LO}t)}_{\text{LO feedthrough}}\\ -{1\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-9373c18d66788a4dbe301f3eb6cb1457_l3.svg "Rendered by QuickLaTeX.com")
Eq.(3) points out that, the output of a single ended mixer contain DC, RF feed-through, LO feed-through and mixed components.
RF feedthough to the output is due to DC component in LO signal.
LO feedthrough is due to DC component in the RF signal.
Figure 2. Output voltage spectrum of single ended switching mixer
Figure 2. illustrates the output voltage spectrum of a single ended switching mixer. The LO feedthrough problem of single ended mixer can be mitigated through single balanced switching mixer using differential LO signals.