How do I default universal reference parameter

I'm writing a function which takes a universal reference and want to give it a default type and value.

This code works.

template <class T = myclass>
void f(T&& t = T{});

Is this more proper way to do it?

template <class T = myclass>
void f(T&& t = typename std::remove_reference<T>::type{});

T will be either one of otherclass&, const otherclass&, otherclass, and it seems that it's not proper to default construct from the reference type.

otherclass o;
const otherclass co;

f(o);            // T will be 'otherclass&'
f(co);           // T will be 'const otherclass&'
f(otherclass{}); // T will be 'otherclass'

I checked if I could construct from the reference type and got erros from this code.

using ref = otherclass&;
otherclass a = ref{};

error: invalid cast of an rvalue expression of type ‘otherclass’ to type ‘ref {aka otherclass&}’

Can anyone help me understand it?