Transfer function of high pass filter via impulse response function
If you consider the simple RC highpass:



Then you can write the two I/O equations:

$$v_i(t)=v_C(t)+v_R(t)$$
$$v_o(t)=v_R(t)$$

Consdering i(t) the current through the circuit (no load) and vC(t) the voltage across C:

$$v_i(t)=\frac{1}{C}\int{i(t)\text{d}t}+R i(t)$$
$$v_o(t)=R i(t)$$

Apply the Laplace transform to the first, I(s) being the Laplace transform of i(t):

$$V_i(s)=\frac{1}{sC}I(s)+R I(s)=I(s)\left(R+\frac{1}{sC}\right)=&gt;$$
$$I(s)=\frac{V_i(s)}{R+\frac{1}{sC}}$$

The same Laplace transform for the output yields:

$$V_o(s)=R I(s)=\frac{V_i(s)R}{R+\frac{1}{sC}}=&gt;$$
$$\frac{V_o(s)}{V_i(s)}=\frac{R}{R+\frac{1}{sC}}=\frac{R}{\frac{sRC+1}{sC}}=\frac{sRC}{sRC+1}=\frac{s}{s+\frac{1}{RC}}$$

This is why I said it should be something like s/(s+1). Now, if you do some inverse Laplace transforms, you'll end up with an interesting impulse response. First, arrange into strictly proper partial fractions:

$$\frac{s}{s+\frac{1}{RC}}=1-\frac{\frac{1}{RC}}{s+\frac{1}{RC}}$$

And now you see that 1 is the Laplace transform of the Dirac impulse, plus the rest of it, which is the lowpass RC with the impulse response \$1-exp(-\frac{t}{RC})\$, and you might be tempted to cancel the 1s, but the first one is the \$\delta\$(t), and the derivative of the step response is \$\frac{exp(-\frac{t}{RC})}{RC}\$, which results in the total impuse response (the point where you shoul have started from):

$$v_o(t)=\delta(t)-\frac{exp(-\frac{t}{RC})}{RC}$$

Here's the confirmation (the input impulse is pulse 0 1k 0 1n 1n 1m -- 1kV over 1ms):



and here's a zoom on the Y axis:



This is one reason why it didn't work out, you omitted the initial conditions and the influence of the Dirac impulse: at t&lt;0 everything is zero (null conditions), at t=0 the (ideal) capacitor charges with the input, the derivative of the applied voltage, \$\delta\$(t), but the input voltage is not just a rise, it's also a fall, both in the same time (Dirac or, how friends called him, Chuck Norris), so the voltage across the capacitor goes back and then reaches its negative peak, after which the discharge happens.